I post multiple choice JavaScript questions on my Instagram stories, which I'll also post here! Last updated: June 12th
From basic to advanced: test how well you know JavaScript, refresh your knowledge a bit, or prepare for your coding interview! ๐ช ๐ I update this repo regularly with new questions. I added the answers in the collapsed sections below the questions, simply click on them to expand it. It's just for fun, good luck! โค๏ธ
Feel free to reach out to me! ๐
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| Feel free to use them in a project! ๐ I would really appreciate a reference to this repo, I create the questions and explanations (yes I'm sad lol) and the community helps me so much to maintain and improve it! ๐ช๐ผ Thank you and have fun! |
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function sayHi() {
console.log(name);
console.log(age);
var name = 'Lydia';
let age = 21;
}
sayHi();- A:
Lydiaandundefined - B:
LydiaandReferenceError - C:
ReferenceErrorand21 - D:
undefinedandReferenceError
Answer
Within the function, we first declare the name variable with the var keyword. This means that the variable gets hoisted (memory space is set up during the creation phase) with the default value of undefined, until we actually get to the line where we define the variable. We haven't defined the variable yet on the line where we try to log the name variable, so it still holds the value of undefined.
Variables with the let keyword (and const) are hoisted, but unlike var, don't get initialized. They are not accessible before the line we declare (initialize) them. This is called the "temporal dead zone". When we try to access the variables before they are declared, JavaScript throws a ReferenceError.
for (var i = 0; i < 3; i++) {
setTimeout(() => console.log(i), 1);
}
for (let i = 0; i < 3; i++) {
setTimeout(() => console.log(i), 1);
}- A:
0 1 2and0 1 2 - B:
0 1 2and3 3 3 - C:
3 3 3and0 1 2
Answer
Because of the event queue in JavaScript, the setTimeout callback function is called after the loop has been executed. Since the variable i in the first loop was declared using the var keyword, this value was global. During the loop, we incremented the value of i by 1 each time, using the unary operator ++. By the time the setTimeout callback function was invoked, i was equal to 3 in the first example.
In the second loop, the variable i was declared using the let keyword: variables declared with the let (and const) keyword are block-scoped (a block is anything between { }). During each iteration, i will have a new value, and each value is scoped inside the loop.
const shape = {
radius: 10,
diameter() {
return this.radius * 2;
},
perimeter: () => 2 * Math.PI * this.radius,
};
console.log(shape.diameter());
console.log(shape.perimeter());- A:
20and62.83185307179586 - B:
20andNaN - C:
20and63 - D:
NaNand63
Answer
Note that the value of diameter is a regular function, whereas the value of perimeter is an arrow function.
With arrow functions, the this keyword refers to its current surrounding scope, unlike regular functions! This means that when we call perimeter, it doesn't refer to the shape object, but to its surrounding scope (window for example).
There is no value radius on that object, which returns NaN.
+true;
!'Lydia';- A:
1andfalse - B:
falseandNaN - C:
falseandfalse
Answer
The unary plus tries to convert an operand to a number. true is 1, and false is 0.
The string 'Lydia' is a truthy value. What we're actually asking, is "is this truthy value falsy?". This returns false.
const bird = {
size: 'small',
};
const mouse = {
name: 'Mickey',
small: true,
};- A:
mouse.bird.sizeis not valid - B:
mouse[bird.size]is not valid - C:
mouse[bird["size"]]is not valid - D: All of them are valid
Answer
In JavaScript, all object keys are strings (unless it's a Symbol). Even though we might not type them as strings, they are always converted into strings under the hood.
JavaScript interprets (or unboxes) statements. When we use bracket notation, it sees the first opening bracket [ and keeps going until it finds the closing bracket ]. Only then, it will evaluate the statement.
mouse[bird.size]: First it evaluates bird.size, which is "small". mouse["small"] returns true
However, with dot notation, this doesn't happen. mouse does not have a key called bird, which means that mouse.bird is undefined. Then, we ask for the size using dot notation: mouse.bird.size. Since mouse.bird is undefined, we're actually asking undefined.size. This isn't valid, and will throw an error similar to Cannot read property "size" of undefined.
let c = { greeting: 'Hey!' };
let d;
d = c;
c.greeting = 'Hello';
console.log(d.greeting);- A:
Hello - B:
Hey! - C:
undefined - D:
ReferenceError - E:
TypeError
Answer
In JavaScript, all objects interact by reference when setting them equal to each other.
First, variable c holds a value to an object. Later, we assign d with the same reference that c has to the object.
When you change one object, you change all of them.
let a = 3;
let b = new Number(3);
let c = 3;
console.log(a == b);
console.log(a === b);
console.log(b === c);- A:
truefalsetrue - B:
falsefalsetrue - C:
truefalsefalse - D:
falsetruetrue
Answer
new Number() is a built-in function constructor. Although it looks like a number, it's not really a number: it has a bunch of extra features and is an object.
When we use the == operator (Equality operator), it only checks whether it has the same value. They both have the value of 3, so it returns true.
However, when we use the === operator (Strict equality operator), both value and type should be the same. It's not: new Number() is not a number, it's an object. Both return false.
class Chameleon {
static colorChange(newColor) {
this.newColor = newColor;
return this.newColor;
}
constructor({ newColor = 'green' } = {}) {
this.newColor = newColor;
}
}
const freddie = new Chameleon({ newColor: 'purple' });
console.log(freddie.colorChange('orange'));- A:
orange - B:
purple - C:
green - D:
TypeError
Answer
The colorChange function is static. Static methods are designed to live only on the constructor in which they are created, and cannot be passed down to any children or called upon class instances. Since freddie is an instance of class Chameleon, the function cannot be called upon it. A TypeError is thrown.
let greeting;
greetign = {}; // Typo!
console.log(greetign);- A:
{} - B:
ReferenceError: greetign is not defined - C:
undefined
Answer
It logs the object, because we just created an empty object on the global object! When we mistyped greeting as greetign, the JS interpreter actually saw this as:
global.greetign = {}in Node.jswindow.greetign = {},frames.greetign = {}andself.greetignin browsers.self.greetignin web workers.globalThis.greetignin all environments.
In order to avoid this, we can use "use strict". This makes sure that you have declared a variable before setting it equal to anything.
function bark() {
console.log('Woof!');
}
bark.animal = 'dog';- A: Nothing, this is totally fine!
- B:
SyntaxError. You cannot add properties to a function this way. - C:
"Woof"gets logged. - D:
ReferenceError
Answer
This is possible in JavaScript, because functions are objects! (Everything besides primitive types are objects)
A function is a special type of object. The code you write yourself isn't the actual function. The function is an object with properties. This property is invocable.
function Person(firstName, lastName) {
this.firstName = firstName;
this.lastName = lastName;
}
const member = new Person('Lydia', 'Hallie');
Person.getFullName = function() {
return `${this.firstName} ${this.lastName}`;
};
console.log(member.getFullName());- A:
TypeError - B:
SyntaxError - C:
Lydia Hallie - D:
undefinedundefined
Answer
In JavaScript, functions are objects, and therefore, the method getFullName gets added to the constructor function object itself. For that reason, we can call Person.getFullName(), but member.getFullName throws a TypeError.
If you want a method to be available to all object instances, you have to add it to the prototype property:
Person.prototype.getFullName = function() {
return `${this.firstName} ${this.lastName}`;
};function Person(firstName, lastName) {
this.firstName = firstName;
this.lastName = lastName;
}
const lydia = new Person('Lydia', 'Hallie');
const sarah = Person('Sarah', 'Smith');
console.log(lydia);
console.log(sarah);- A:
Person {firstName: "Lydia", lastName: "Hallie"}andundefined - B:
Person {firstName: "Lydia", lastName: "Hallie"}andPerson {firstName: "Sarah", lastName: "Smith"} - C:
Person {firstName: "Lydia", lastName: "Hallie"}and{} - D:
Person {firstName: "Lydia", lastName: "Hallie"}andReferenceError
Answer
For sarah, we didn't use the new keyword. When using new, this refers to the new empty object we create. However, if you don't add new, this refers to the global object!
We said that this.firstName equals "Sarah" and this.lastName equals "Smith". What we actually did, is defining global.firstName = 'Sarah' and global.lastName = 'Smith'. sarah itself is left undefined, since we don't return a value from the Person function.
- A: Target > Capturing > Bubbling
- B: Bubbling > Target > Capturing
- C: Target > Bubbling > Capturing
- D: Capturing > Target > Bubbling
Answer
During the capturing phase, the event goes through the ancestor elements down to the target element. It then reaches the target element, and bubbling begins.
- A: true
- B: false
Answer
All objects have prototypes, except for the base object. The base object is the object created by the user, or an object that is created using the new keyword. The base object has access to some methods and properties, such as .toString. This is the reason why you can use built-in JavaScript methods! All of such methods are available on the prototype. Although JavaScript can't find it directly on your object, it goes down the prototype chain and finds it there, which makes it accessible for you.
function sum(a, b) {
return a + b;
}
sum(1, '2');- A:
NaN - B:
TypeError - C:
"12" - D:
3
Answer
JavaScript is a dynamically typed language: we don't specify what types certain variables are. Values can automatically be converted into another type without you knowing, which is called implicit type coercion. Coercion is converting from one type into another.
In this example, JavaScript converts the number 1 into a string, in order for the function to make sense and return a value. During the addition of a numeric type (1) and a string type ('2'), the number is treated as a string. We can concatenate strings like "Hello" + "World", so what's happening here is "1" + "2" which returns "12".
let number = 0;
console.log(number++);
console.log(++number);
console.log(number);- A:
112 - B:
122 - C:
022 - D:
012
Answer
The postfix unary operator ++:
- Returns the value (this returns
0) - Increments the value (number is now
1)
The prefix unary operator ++:
- Increments the value (number is now
2) - Returns the value (this returns
2)
This returns 0 2 2.
function getPersonInfo(one, two, three) {
console.log(one);
console.log(two);
console.log(three);
}
const person = 'Lydia';
const age = 21;
getPersonInfo`${person} is ${age} years old`;- A:
"Lydia"21["", " is ", " years old"] - B:
["", " is ", " years old"]"Lydia"21 - C:
"Lydia"["", " is ", " years old"]21
Answer
If you use tagged template literals, the value of the first argument is always an array of the string values. The remaining arguments get the values of the passed expressions!
function checkAge(data) {
if (data === { age: 18 }) {
console.log('You are an adult!');
} else if (data == { age: 18 }) {
console.log('You are still an adult.');
} else {
console.log(`Hmm.. You don't have an age I guess`);
}
}
checkAge({ age: 18 });- A:
You are an adult! - B:
You are still an adult. - C:
Hmm.. You don't have an age I guess
Answer
When testing equality, primitives are compared by their value, while objects are compared by their reference. JavaScript checks if the objects have a reference to the same location in memory.
The two objects that we are comparing don't have that: the object we passed as a parameter refers to a different location in memory than the object we used in order to check equality.
This is why both { age: 18 } === { age: 18 } and { age: 18 } == { age: 18 } return false.
function getAge(...args) {
console.log(typeof args);
}
getAge(21);- A:
"number" - B:
"array" - C:
"object" - D:
"NaN"
Answer
The rest parameter (...args) lets us "collect" all remaining arguments into an array. An array is an object, so typeof args returns "object"
function getAge() {
'use strict';
age = 21;
console.log(age);
}
getAge();- A:
21 - B:
undefined - C:
ReferenceError - D:
TypeError
Answer
With "use strict", you can make sure that you don't accidentally declare global variables. We never declared the variable age, and since we use "use strict", it will throw a reference error. If we didn't use "use strict", it would have worked, since the property age would have gotten added to the global object.
const sum = eval('10*10+5');- A:
105 - B:
"105" - C:
TypeError - D:
"10*10+5"
Answer
eval evaluates codes that's passed as a string. If it's an expression, like in this case, it evaluates the expression. The expression is 10 * 10 + 5. This returns the number 105.
sessionStorage.setItem('cool_secret', 123);- A: Forever, the data doesn't get lost.
- B: When the user closes the tab.
- C: When the user closes the entire browser, not only the tab.
- D: When the user shuts off their computer.
Answer
The data stored in sessionStorage is removed after closing the tab.
If you used localStorage, the data would've been there forever, unless for example localStorage.clear() is invoked.
var num = 8;
var num = 10;
console.log(num);- A:
8 - B:
10 - C:
SyntaxError - D:
ReferenceError
Answer
With the var keyword, you can declare multiple variables with the same name. The variable will then hold the latest value.
You cannot do this with let or const since they're block-scoped.
const obj = { 1: 'a', 2: 'b', 3: 'c' };
const set = new Set([1, 2, 3, 4, 5]);
obj.hasOwnProperty('1');
obj.hasOwnProperty(1);
set.has('1');
set.has(1);- A:
falsetruefalsetrue - B:
falsetruetruetrue - C:
truetruefalsetrue - D:
truetruetruetrue
Answer
All object keys (excluding Symbols) are strings under the hood, even if you don't type it yourself as a string. This is why obj.hasOwnProperty('1') also returns true.
It doesn't work that way for a set. There is no '1' in our set: set.has('1') returns false. It has the numeric type 1, set.has(1) returns true.
const obj = { a: 'one', b: 'two', a: 'three' };
console.log(obj);- A:
{ a: "one", b: "two" } - B:
{ b: "two", a: "three" } - C:
{ a: "three", b: "two" } - D:
SyntaxError
Answer
If you have two keys with the same name, the key will be replaced. It will still be in its first position, but with the last specified value.
26. The JavaScript global execution context creates two things for you: the global object, and the "this" keyword.
- A: true
- B: false
- C: it depends
Answer
The base execution context is the global execution context: it's what's accessible everywhere in your code.
for (let i = 1; i < 5; i++) {
if (i === 3) continue;
console.log(i);
}- A:
12 - B:
123 - C:
124 - D:
134
String.prototype.giveLydiaPizza = () => {
return 'Just give Lydia pizza already!';
};
const name = 'Lydia';
console.log(name.giveLydiaPizza())- A:
"Just give Lydia pizza already!" - B:
TypeError: not a function - C:
SyntaxError - D:
undefined
Answer
String is a built-in constructor, which we can add properties to. I just added a method to its prototype. Primitive strings are automatically converted into a string object, generated by the string prototype function. So, all strings (string objects) have access to that method!
const a = {};
const b = { key: 'b' };
const c = { key: 'c' };
a[b] = 123;
a[c] = 456;
console.log(a[b]);- A:
123 - B:
456 - C:
undefined - D:
ReferenceError
Answer
Object keys are automatically converted into strings. We are trying to set an object as a key to object a, with the value of 123.
However, when we stringify an object, it becomes "[object Object]". So what we are saying here, is that a["[object Object]"] = 123. Then, we can try to do the same again. c is another object that we are implicitly stringifying. So then, a["[object Object]"] = 456.
Then, we log a[b], which is actually a["[object Object]"]. We just set that to 456, so it returns 456.
const foo = () => console.log('First');
const bar = () => setTimeout(() => console.log('Second'));
const baz = () => console.log('Third');
bar();
foo();
baz();- A:
FirstSecondThird - B:
FirstThirdSecond - C:
SecondFirstThird - D:
SecondThirdFirst
Answer
We have a setTimeout function and invoked it first. Yet, it was logged last.
This is because in browsers, we don't just have the runtime engine, we also have something called a WebAPI. The WebAPI gives us the setTimeout function to start with, and for example the DOM.
After the callback is pushed to the WebAPI, the setTimeout function itself (but not the callback!) is popped off the stack.
Now, foo gets invoked, and "First" is being logged.
foo is popped off the stack, and baz gets invoked. "Third" gets logged.
The WebAPI can't just add stuff to the stack whenever it's ready. Instead, it pushes the callback function to something called the queue.
This is where an event loop starts to work. An event loop looks at the stack and task queue. If the stack is empty, it takes the first thing on the queue and pushes it onto the stack.
bar gets invoked, "Second" gets logged, and it's popped off the stack.
<div onclick="console.log('first div')">
<div onclick="console.log('second div')">
<button onclick="console.log('button')">
Click!
</button>
</div>
</div>- A: Outer
div - B: Inner
div - C:
button - D: An array of all nested elements.
Answer
The deepest nested element that caused the event is the target of the event. You can stop bubbling by event.stopPropagation
<div onclick="console.log('div')">
<p onclick="console.log('p')">
Click here!
</p>
</div>- A:
pdiv - B:
divp - C:
p - D:
div
Answer
If we click p, we see two logs: p and div. During event propagation, there are 3 phases: capturing, target, and bubbling. By default, event handlers are executed in the bubbling phase (unless you set useCapture to true). It goes from the deepest nested element outwards.
const person = { name: 'Lydia' };
function sayHi(age) {
return `${this.name} is ${age}`;
}
console.log(sayHi.call(person, 21));
console.log(sayHi.bind(person, 21));- A:
undefined is 21Lydia is 21 - B:
functionfunction - C:
Lydia is 21Lydia is 21 - D:
Lydia is 21function
Answer
With both, we can pass the object to which we want the this keyword to refer to. However, .call is also executed immediately!
.bind. returns a copy of the function, but with a bound context! It is not executed immediately.
function sayHi() {
return (() => 0)();
}
console.log(typeof sayHi());- A:
"object" - B:
"number" - C:
"function" - D:
"undefined"
Answer
The sayHi function returns the returned value of the immediately invoked function expression (IIFE). This function returned 0, which is type "number".
FYI: typeof can return the following list of values: undefined, boolean, number, bigint, string, symbol, function and object. Note that typeof null returns "object".
0;
new Number(0);
('');
(' ');
new Boolean(false);
undefined;- A:
0,'',undefined - B:
0,new Number(0),'',new Boolean(false),undefined - C:
0,'',new Boolean(false),undefined - D: All of them are falsy
Answer
There are 8 falsy values:
undefinednullNaNfalse''(empty string)0-00n(BigInt(0))
Function constructors, like new Number and new Boolean are truthy.
console.log(typeof typeof 1);- A:
"number" - B:
"string" - C:
"object" - D:
"undefined"
const numbers = [1, 2, 3];
numbers[10] = 11;
console.log(numbers);- A:
[1, 2, 3, null x 7, 11] - B:
[1, 2, 3, 11] - C:
[1, 2, 3, empty x 7, 11] - D:
SyntaxError
Answer
When you set a value to an element in an array that exceeds the length of the array, JavaScript creates something called "empty slots". These actually have the value of undefined, but you will see something like:
[1, 2, 3, empty x 7, 11]
depending on where you run it (it's different for every browser, node, etc.)
(() => {
let x, y;
try {
throw new Error();
} catch (x) {
(x = 1), (y = 2);
console.log(x);
}
console.log(x);
console.log(y);
})();- A:
1undefined2 - B:
undefinedundefinedundefined - C:
112 - D:
1undefinedundefined
Answer
The catch block receives the argument x. This is not the same x as the variable when we pass arguments. This variable x is block-scoped.
Later, we set this block-scoped variable equal to 1, and set the value of the variable y. Now, we log the block-scoped variable x, which is equal to 1.
Outside of the catch block, x is still undefined, and y is 2. When we want to console.log(x) outside of the catch block, it returns undefined, and y returns 2.
- A: primitive or object
- B: function or object
- C: trick question! only objects
- D: number or object
Answer
JavaScript only has primitive types and objects.
Primitive types are boolean, null, undefined, bigint, number, string, and symbol.
What differentiates a primitive from an object is that primitives do not have any properties or methods; however, you'll note that 'foo'.toUpperCase() evaluates to 'FOO' and does not result in a TypeError. This is because when you try to access a property or method on a primitive like a string, JavaScript will implicitly wrap the primitive type using one of the wrapper classes, i.e. String, and then immediately discard the wrapper after the expression evaluates. All primitives except for null and undefined exhibit this behaviour.
[[0, 1], [2, 3]].reduce(
(acc, cur) => {
return acc.concat(cur);
},
[1, 2],
);- A:
[0, 1, 2, 3, 1, 2] - B:
[6, 1, 2] - C:
[1, 2, 0, 1, 2, 3] - D:
[1, 2, 6]
Answer
[1, 2] is our initial value. This is the value we start with, and the value of the very first acc. During the first round, acc is [1,2], and cur is [0, 1]. We concatenate them, which results in [1, 2, 0, 1].
Then, [1, 2, 0, 1] is acc and [2, 3] is cur. We concatenate them, and get [1, 2, 0, 1, 2, 3]
!!null;
!!'';
!!1;- A:
falsetruefalse - B:
falsefalsetrue - C:
falsetruetrue - D:
truetruefalse
Answer
null is falsy. !null returns true. !true returns false.
"" is falsy. !"" returns true. !true returns false.
1 is truthy. !1 returns false. !false returns true.
setInterval(() => console.log('Hi'), 1000);- A: a unique id
- B: the amount of milliseconds specified
- C: the passed function
- D:
undefined
Answer
It returns a unique id. This id can be used to clear that interval with the clearInterval() function.
[...'Lydia'];- A:
["L", "y", "d", "i", "a"] - B:
["Lydia"] - C:
[[], "Lydia"] - D:
[["L", "y", "d", "i", "a"]]
Answer
A string is an iterable. The spread operator maps every character of an iterable to one element.
function* generator(i) {
yield i;
yield i * 2;
}
const gen = generator(10);
console.log(gen.next().value);
console.log(gen.next().value);- A:
[0, 10], [10, 20] - B:
20, 20 - C:
10, 20 - D:
0, 10 and 10, 20
Answer
Regular functions cannot be stopped mid-way after invocation. However, a generator function can be "stopped" midway, and later continue from where it stopped. Every time a generator function encounters a yield keyword, the function yields the value specified after it. Note that the generator function in that case doesnโt return the value, it yields the value.
First, we initialize the generator function with i equal to 10. We invoke the generator function using the next() method. The first time we invoke the generator function, i is equal to 10. It encounters the first yield keyword: it yields the value of i. The generator is now "paused", and 10 gets logged.
Then, we invoke the function again with the next() method. It starts to continue where it stopped previously, still with i equal to 10. Now, it encounters the next yield keyword, and yields i * 2. i is equal to 10, so it returns 10 * 2, which is 20. This results in 10, 20.
const firstPromise = new Promise((res, rej) => {
setTimeout(res, 500, 'one');
});
const secondPromise = new Promise((res, rej) => {
setTimeout(res, 100, 'two');
});
Promise.race([firstPromise, secondPromise]).then(res => console.log(res));- A:
"one" - B:
"two" - C:
"two" "one" - D:
"one" "two"
Answer
When we pass multiple promises to the Promise.race method, it resolves/rejects the first promise that resolves/rejects. To the setTimeout method, we pass a timer: 500ms for the first promise (firstPromise), and 100ms for the second promise (secondPromise). This means that the secondPromise resolves first with the value of 'two'. res now holds the value of 'two', which gets logged.
let person = { name: 'Lydia' };
const members = [person];
person = null;
console.log(members);- A:
null - B:
[null] - C:
[{}] - D:
[{ name: "Lydia" }]
Answer
First, we declare a variable person with the value of an object that has a name property.
Then, we declare a variable called members. We set the first element of that array equal to the value of the person variable. Objects interact by reference when setting them equal to each other. When you assign a reference from one variable to another, you make a copy of that reference. (note that they don't have the same reference!)
Then, we set the variable person equal to null.
We are only modifying the value of the person variable, and not the first element in the array, since that element has a different (copied) reference to the object. The first element in members still holds its reference to the original object. When we log the members array, the first element still holds the value of the object, which gets logged.
const person = {
name: 'Lydia',
age: 21,
};
for (const item in person) {
console.log(item);
}- A:
{ name: "Lydia" }, { age: 21 } - B:
"name", "age" - C:
"Lydia", 21 - D:
["name", "Lydia"], ["age", 21]
Answer
With a for-in loop, we can iterate through object keys, in this case name and age. Under the hood, object keys are strings (if they're not a Symbol). On every loop, we set the value of item equal to the current key itโs iterating over. First, item is equal to name, and gets logged. Then, item is equal to age, which gets logged.
console.log(3 + 4 + '5');- A:
"345" - B:
"75" - C:
12 - D:
"12"
Answer
Operator associativity is the order in which the compiler evaluates the expressions, either left-to-right or right-to-left. This only happens if all operators have the same precedence. We only have one type of operator: +. For addition, the associativity is left-to-right.
3 + 4 gets evaluated first. This results in the number 7.
7 + '5' results in "75" because of coercion. JavaScript converts the number 7 into a string, see question 15. We can concatenate two strings using the +operator. "7" + "5" results in "75".
const num = parseInt('7*6', 10);- A:
42 - B:
"42" - C:
7 - D:
NaN
Answer
Only the first numbers in the string is returned. Based on the radix (the second argument in order to specify what type of number we want to parse it to: base 10, hexadecimal, octal, binary, etc.), the parseInt checks whether the characters in the string are valid. Once it encounters a character that isn't a valid number in the radix, it stops parsing and ignores the following characters.
* is not a valid number. It only parses "7" into the decimal 7. num now holds the value of 7.
[1, 2, 3].map(num => {
if (typeof num === 'number') return;
return num * 2;
});- A:
[] - B:
[null, null, null] - C:
[undefined, undefined, undefined] - D:
[ 3 x empty ]
Answer
When mapping over the array, the value of num is equal to the element itโs currently looping over. In this case, the elements are numbers, so the condition of the if statement typeof num === "number" returns true. The map function creates a new array and inserts the values returned from the function.
However, we donโt return a value. When we donโt return a value from the function, the function returns undefined. For every element in the array, the function block gets called, so for each element we return undefined.
function getInfo(member, year) {
member.name = 'Lydia';
year = '1998';
}
const person = { name: 'Sarah' };
const birthYear = '1997';
getInfo(person, birthYear);
console.log(person, birthYear);- A:
{ name: "Lydia" }, "1997" - B:
{ name: "Sarah" }, "1998" - C:
{ name: "Lydia" }, "1998" - D:
{ name: "Sarah" }, "1997"
Answer
Arguments are passed by value, unless their value is an object, then they're passed by reference. birthYear is passed by value, since it's a string, not an object. When we pass arguments by value, a copy of that value is created (see question 46).
The variable birthYear has a reference to the value "1997". The argument year also has a reference to the value "1997", but it's not the same value as birthYear has a reference to. When we update the value of year by setting year equal to "1998", we are only updating the value of year. birthYear is still equal to "1997".
The value of person is an object. The argument member has a (copied) reference to the same object. When we modify a property of the object member has a reference to, the value of person will also be modified, since they both have a reference to the same object. person's name property is now equal to the value "Lydia"
function greeting() {
throw 'Hello world!';
}
function sayHi() {
try {
const data = greeting();
console.log('It worked!', data);
} catch (e) {
console.log('Oh no an error:', e);
}
}
sayHi();- A:
It worked! Hello world! - B:
Oh no an error: undefined - C:
SyntaxError: can only throw Error objects - D:
Oh no an error: Hello world!
Answer
With the throw statement, we can create custom errors. With this statement, you can throw exceptions. An exception can be a string, a number, a boolean or an object. In this case, our exception is the string 'Hello world!'.
With the catch statement, we can specify what to do if an exception is thrown in the try block. An exception is thrown: the string 'Hello world!'. e is now equal to that string, which we log. This results in 'Oh an error: Hello world!'.
function Car() {
this.make = 'Lamborghini';
return { make: 'Maserati' };
}
const myCar = new Car();
console.log(myCar.make);- A:
"Lamborghini" - B:
"Maserati" - C:
ReferenceError - D:
TypeError
Answer
When you return a property, the value of the property is equal to the returned value, not the value set in the constructor function. We return the string "Maserati", so myCar.make is equal to "Maserati".
(() => {
let x = (y = 10);
})();
console.log(typeof x);
console.log(typeof y);- A:
"undefined", "number" - B:
"number", "number" - C:
"object", "number" - D:
"number", "undefined"
Answer
let x = (y = 10); is actually shorthand for:
y = 10;
let x = y;When we set y equal to 10, we actually add a property y to the global object (window in browser, global in Node). In a browser, window.y is now equal to 10.
Then, we declare a variable x with the value of y, which is 10. Variables declared with the let keyword are block scoped, they are only defined within the block they're declared in; the immediately invoked function expression (IIFE) in this case. When we use the typeof operator, the operand x is not defined: we are trying to access x outside of the block it's declared in. This means that x is not defined. Values who haven't been assigned a value or declared are of type "undefined". console.log(typeof x) returns "undefined".
However, we created a global variable y when setting y equal to 10. This value is accessible anywhere in our code. y is defined, and holds a value of type "number". console.log(typeof y) returns "number".
class Dog {
constructor(name) {
this.name = name;
}
}
Dog.prototype.bark = function() {
console.log(`Woof I am ${this.name}`);
};
const pet = new Dog('Mara');
pet.bark();
delete Dog.prototype.bark;
pet.bark();- A:
"Woof I am Mara",TypeError - B:
"Woof I am Mara","Woof I am Mara" - C:
"Woof I am Mara",undefined - D:
TypeError,TypeError
Answer
We can delete properties from objects using the delete keyword, also on the prototype. By deleting a property on the prototype, it is not available anymore in the prototype chain. In this case, the bark function is not available anymore on the prototype after delete Dog.prototype.bark, yet we still try to access it.
When we try to invoke something that is not a function, a TypeError is thrown. In this case TypeError: pet.bark is not a function, since pet.bark is undefined.
const set = new Set([1, 1, 2, 3, 4]);
console.log(set);- A:
[1, 1, 2, 3, 4] - B:
[1, 2, 3, 4] - C:
{1, 1, 2, 3, 4} - D:
{1, 2, 3, 4}
Answer
The Set object is a collection of unique values: a value can only occur once in a set.
We passed the iterable [1, 1, 2, 3, 4] with a duplicate value 1. Since we cannot have two of the same values in a set, one of them is removed. This results in {1, 2, 3, 4}.
// counter.js
let counter = 10;
export default counter;// index.js
import myCounter from './counter';
myCounter += 1;
console.log(myCounter);- A:
10 - B:
11 - C:
Error - D:
NaN
Answer
An imported module is read-only: you cannot modify the imported module. Only the module that exports them can change its value.
When we try to increment the value of myCounter, it throws an error: myCounter is read-only and cannot be modified.
const name = 'Lydia';
age = 21;
console.log(delete name);
console.log(delete age);- A:
false,true - B:
"Lydia",21 - C:
true,true - D:
undefined,undefined
Answer
The delete operator returns a boolean value: true on a successful deletion, else it'll return false. However, variables declared with the var, const or let keyword cannot be deleted using the delete operator.
The name variable was declared with a const keyword, so its deletion is not successful: false is returned. When we set age equal to 21, we actually added a property called age to the global object. You can successfully delete properties from objects this way, also the global object, so delete age returns true.
const numbers = [1, 2, 3, 4, 5];
const [y] = numbers;
console.log(y);- A:
[[1, 2, 3, 4, 5]] - B:
[1, 2, 3, 4, 5] - C:
1 - D:
[1]
Answer
We can unpack values from arrays or properties from objects through destructuring. For example:
[a, b] = [1, 2];
The value of a is now 1, and the value of b is now 2. What we actually did in the question, is:
[y] = [1, 2, 3, 4, 5];
This means that the value of y is equal to the first value in the array, which is the number 1. When we log y, 1 is returned.
const user = { name: 'Lydia', age: 21 };
const admin = { admin: true, ...user };
console.log(admin);- A:
{ admin: true, user: { name: "Lydia", age: 21 } } - B:
{ admin: true, name: "Lydia", age: 21 } - C:
{ admin: true, user: ["Lydia", 21] } - D:
{ admin: true }
Answer
It's possible to combine objects using the spread operator .... It lets you create copies of the key/value pairs of one object, and add them to another object. In this case, we create copies of the user object, and add them to the admin object. The admin object now contains the copied key/value pairs, which results in { admin: true, name: "Lydia", age: 21 }.
const person = { name: 'Lydia' };
Object.defineProperty(person, 'age', { value: 21 });
console.log(person);
console.log(Object.keys(person));- A:
{ name: "Lydia", age: 21 },["name", "age"] - B:
{ name: "Lydia", age: 21 },["name"] - C:
{ name: "Lydia"},["name", "age"] - D:
{ name: "Lydia"},["age"]
Answer
With the defineProperty method, we can add new properties to an object, or modify existing ones. When we add a property to an object using the defineProperty method, they are by default not enumerable. The Object.keys method returns all enumerable property names from an object, in this case only "name".
Properties added using the defineProperty method are immutable by default. You can override this behavior using the writable, configurable and enumerable properties. This way, the defineProperty method gives you a lot more control over the properties you're adding to an object.
const settings = {
username: 'lydiahallie',
level: 19,
health: 90,
};
const data = JSON.stringify(settings, ['level', 'health']);
console.log(data);- A:
"{"level":19, "health":90}" - B:
"{"username": "lydiahallie"}" - C:
"["level", "health"]" - D:
"{"username": "lydiahallie", "level":19, "health":90}"
Answer
The second argument of JSON.stringify is the replacer. The replacer can either be a function or an array, and lets you control what and how the values should be stringified.
If the replacer is an array, only the property names included in the array will be added to the JSON string. In this case, only the properties with the names "level" and "health" are included, "username" is excluded. data is now equal to "{"level":19, "health":90}".
If the replacer is a function, this function gets called on every property in the object you're stringifying. The value returned from this function will be the value of the property when it's added to the JSON string. If the value is undefined, this property is excluded from the JSON string.
let num = 10;
const increaseNumber = () => num++;
const increasePassedNumber = number => number++;
const num1 = increaseNumber();
const num2 = increasePassedNumber(num1);
console.log(num1);
console.log(num2);- A:
10,10 - B:
10,11 - C:
11,11 - D:
11,12
Answer
The unary operator ++ first returns the value of the operand, then increments the value of the operand. The value of num1 is 10, since the increaseNumber function first returns the value of num, which is 10, and only increments the value of num afterwards.
num2 is 10, since we passed num1 to the increasePassedNumber. number is equal to 10(the value of num1. Again, the unary operator ++ first returns the value of the operand, then increments the value of the operand. The value of number is 10, so num2 is equal to 10.
const value = { number: 10 };
const multiply = (x = { ...value }) => {
console.log((x.number *= 2));
};
multiply();
multiply();
multiply(value);
multiply(value);- A:
20,40,80,160 - B:
20,40,20,40 - C:
20,20,20,40 - D:
NaN,NaN,20,40
Answer
In ES6, we can initialize parameters with a default value. The value of the parameter will be the default value, if no other value has been passed to the function, or if the value of the parameter is "undefined". In this case, we spread the properties of the value object into a new object, so x has the default value of { number: 10 }.
The default argument is evaluated at call time! Every time we call the function, a new object is created. We invoke the multiply function the first two times without passing a value: x has the default value of { number: 10 }. We then log the multiplied value of that number, which is 20.
The third time we invoke multiply, we do pass an argument: the object called value. The *= operator is actually shorthand for x.number = x.number * 2: we modify the value of x.number, and log the multiplied value 20.
The fourth time, we pass the value object again. x.number was previously modified to 20, so x.number *= 2 logs 40.
[1, 2, 3, 4].reduce((x, y) => console.log(x, y));- A:
12and33and64 - B:
12and23and34 - C:
1undefinedand2undefinedand3undefinedand4undefined - D:
12andundefined3andundefined4
Answer
The first argument that the reduce method receives is the accumulator, x in this case. The second argument is the current value, y. With the reduce method, we execute a callback function on every element in the array, which could ultimately result in one single value.
In this example, we are not returning any values, we are simply logging the values of the accumulator and the current value.
The value of the accumulator is equal to the previously returned value of the callback function. If you don't pass the optional initialValue argument to the reduce method, the accumulator is equal to the first element on the first call.
On the first call, the accumulator (x) is 1, and the current value (y) is 2. We don't return from the callback function, we log the accumulator and current value: 1 and 2 get logged.
If you don't return a value from a function, it returns undefined. On the next call, the accumulator is undefined, and the current value is 3. undefined and 3 get logged.
On the fourth call, we again don't return from the callback function. The accumulator is again undefined, and the current value is 4. undefined and 4 get logged.
class Dog {
constructor(name) {
this.name = name;
}
};
class Labrador extends Dog {
// 1
constructor(name, size) {
this.size = size;
}
// 2
constructor(name, size) {
super(name);
this.size = size;
}
// 3
constructor(size) {
super(name);
this.size = size;
}
// 4
constructor(name, size) {
this.name = name;
this.size = size;
}
};- A: 1
- B: 2
- C: 3
- D: 4
Answer
In a derived class, you cannot access the this keyword before calling super. If you try to do that, it will throw a ReferenceError: 1 and 4 would throw a reference error.
With the super keyword, we call that parent class's constructor with the given arguments. The parent's constructor receives the name argument, so we need to pass name to super.
The Labrador class receives two arguments, name since it extends Dog, and size as an extra property on the Labrador class. They both need to be passed to the constructor function on Labrador, which is done correctly using constructor 2.
// index.js
console.log('running index.js');
import { sum } from './sum.js';
console.log(sum(1, 2));
// sum.js
console.log('running sum.js');
export const sum = (a, b) => a + b;- A:
running index.js,running sum.js,3 - B:
running sum.js,running index.js,3 - C:
running sum.js,3,running index.js - D:
running index.js,undefined,running sum.js
Answer
With the import keyword, all imported modules are pre-parsed. This means that the imported modules get run first, the code in the file which imports the module gets executed after.
This is a difference between require() in CommonJS and import! With require(), you can load dependencies on demand while the code is being run. If we would have used require instead of import, running index.js, running sum.js, 3 would have been logged to the console.
console.log(Number(2) === Number(2));
console.log(Boolean(false) === Boolean(false));
console.log(Symbol('foo') === Symbol('foo'));- A:
true,true,false - B:
false,true,false - C:
true,false,true - D:
true,true,true
Answer
Every Symbol is entirely unique. The purpose of the argument passed to the Symbol is to give the Symbol a description. The value of the Symbol is not dependent on the passed argument. As we test equality, we are creating two entirely new symbols: the first Symbol('foo'), and the second Symbol('foo'). These two values are unique and not equal to each other, Symbol('foo') === Symbol('foo') returns false.
const name = 'Lydia Hallie';
console.log(name.padStart(13));
console.log(name.padStart(2));- A:
"Lydia Hallie","Lydia Hallie" - B:
" Lydia Hallie"," Lydia Hallie"("[13x whitespace]Lydia Hallie","[2x whitespace]Lydia Hallie") - C:
" Lydia Hallie","Lydia Hallie"("[1x whitespace]Lydia Hallie","Lydia Hallie") - D:
"Lydia Hallie","Lyd",
Answer
With the padStart method, we can add padding to the beginning of a string. The value passed to this method is the total length of the string together with the padding. The string "Lydia Hallie" has a length of 12. name.padStart(13) inserts 1 space at the start of the string, because 12 + 1 is 13.
If the argument passed to the padStart method is smaller than the length of the array, no padding will be added.
console.log('๐ฅ' + '๐ป');- A:
"๐ฅ๐ป" - B:
257548 - C: A string containing their code points
- D: Error
Answer
With the + operator, you can concatenate strings. In this case, we are concatenating the string "๐ฅ" with the string "๐ป", resulting in "๐ฅ๐ป".
function* startGame() {
const answer = yield 'Do you love JavaScript?';
if (answer !== 'Yes') {
return "Oh wow... Guess we're done here";
}
return 'JavaScript loves you back โค๏ธ';
}
const game = startGame();
console.log(/* 1 */); // Do you love JavaScript?
console.log(/* 2 */); // JavaScript loves you back โค๏ธ- A:
game.next("Yes").valueandgame.next().value - B:
game.next.value("Yes")andgame.next.value() - C:
game.next().valueandgame.next("Yes").value - D:
game.next.value()andgame.next.value("Yes")
Answer
A generator function "pauses" its execution when it sees the yield keyword. First, we have to let the function yield the string "Do you love JavaScript?", which can be done by calling game.next().value.
Every line is executed, until it finds the first yield keyword. There is a yield keyword on the first line within the function: the execution stops with the first yield! This means that the variable answer is not defined yet!
When we call game.next("Yes").value, the previous yield is replaced with the value of the parameters passed to the next() function, "Yes" in this case. The value of the variable answer is now equal to "Yes". The condition of the if-statement returns false, and JavaScript loves you back โค๏ธ gets logged.
console.log(String.raw`Hello\nworld`);- A:
Hello world! - B:
Hello
world - C:
Hello\nworld - D:
Hello\n
world
Answer
String.raw returns a string where the escapes (\n, \v, \t etc.) are ignored! Backslashes can be an issue since you could end up with something like:
const path = `C:\Documents\Projects\table.html`
Which would result in:
"C:DocumentsProjects able.html"
With String.raw, it would simply ignore the escape and print:
C:\Documents\Projects\table.html
In this case, the string is Hello\nworld, which gets logged.
async function getData() {
return await Promise.resolve('I made it!');
}
const data = getData();
console.log(data);- A:
"I made it!" - B:
Promise {<resolved>: "I made it!"} - C:
Promise {<pending>} - D:
undefined
Answer
An async function always returns a promise. The await still has to wait for the promise to resolve: a pending promise gets returned when we call getData() in order to set data equal to it.
If we wanted to get access to the resolved value "I made it", we could have used the .then() method on data:
data.then(res => console.log(res))
This would've logged "I made it!"
function addToList(item, list) {
return list.push(item);
}
const result = addToList('apple', ['banana']);
console.log(result);- A:
['apple', 'banana'] - B:
2 - C:
true - D:
undefined
Answer
The .push() method returns the length of the new array! Previously, the array contained one element (the string "banana") and had a length of 1. After adding the string "apple" to the array, the array contains two elements, and has a length of 2. This gets returned from the addToList function.
The push method modifies the original array. If you wanted to return the array from the function rather than the length of the array, you should have returned list after pushing item to it.
const box = { x: 10, y: 20 };
Object.freeze(box);
const shape = box;
shape.x = 100;
console.log(shape);- A:
{ x: 100, y: 20 } - B:
{ x: 10, y: 20 } - C:
{ x: 100 } - D:
ReferenceError
Answer
Object.freeze makes it impossible to add, remove, or modify properties of an object (unless the property's value is another object).
When we create the variable shape and set it equal to the frozen object box, shape also refers to a frozen object. You can check whether an object is frozen by using Object.isFrozen. In this case, Object.isFrozen(shape) would return true, since the variable shape has a reference to a frozen object.
Since shape is frozen, and since the value of x is not an object, we cannot modify the property x. x is still equal to 10, and { x: 10, y: 20 } gets logged.
const { firstName: myName } = { firstName: 'Lydia' };
console.log(firstName);- A:
"Lydia" - B:
"myName" - C:
undefined - D:
ReferenceError
Answer
By using destructuring assignment syntax we can unpack values from arrays, or properties from objects, into distinct variables:
const { firstName } = { firstName: 'Lydia' };
// ES5 version:
// var firstName = { firstName: 'Lydia' }.firstName;
console.log(firstName); // "Lydia"Also, a property can be unpacked from an object and assigned to a variable with a different name than the object property:
const { firstName: myName } = { firstName: 'Lydia' };
// ES5 version:
// var myName = { firstName: 'Lydia' }.firstName;
console.log(myName); // "Lydia"
console.log(firstName); // Uncaught ReferenceError: firstName is not definedTherefore, firstName does not exist as a variable, thus attempting to access its value will raise a ReferenceError.
Note: Be aware of the global scope properties:
const { name: myName } = { name: 'Lydia' };
console.log(myName); // "lydia"
console.log(name); // "" ----- Browser e.g. Chrome
console.log(name); // ReferenceError: name is not defined ----- NodeJSWhenever Javascript is unable to find a variable within the current scope, it climbs up the Scope chain and searches for it and if it reaches the top-level scope, aka Global scope, and still doesn't find it, it will throw a ReferenceError.
-
In Browsers such as Chrome,
nameis a deprecated global scope property. In this example, the code is running inside global scope and there is no user defined local variable forname, therefore it searches the predefined variables/properties in the global scope which is in case of browsers, it searches throughwindowobject and it will extract the window.name value which is equal to an empty string. -
In NodeJS, there is no such property on the
globalobject, thus attempting to access a non-existent variable will raise a ReferenceError.
function sum(a, b) {
return a + b;
}- A: Yes
- B: No
Answer
A pure function is a function that always returns the same result, if the same arguments are passed.
The sum function always returns the same result. If we pass 1 and 2, it will always return 3 without side effects. If we pass 5 and 10, it will always return 15, and so on. This is the definition of a pure function.
const add = () => {
const cache = {};
return num => {
if (num in cache) {
return `From cache! ${cache[num]}`;
} else {
const result = num + 10;
cache[num] = result;
return `Calculated! ${result}`;
}
};
};
const addFunction = add();
console.log(addFunction(10));
console.log(addFunction(10));
console.log(addFunction(5 * 2));- A:
Calculated! 20Calculated! 20Calculated! 20 - B:
Calculated! 20From cache! 20Calculated! 20 - C:
Calculated! 20From cache! 20From cache! 20 - D:
Calculated! 20From cache! 20Error
Answer
The add function is a memoized function. With memoization, we can cache the results of a function in order to speed up its execution. In this case, we create a cache object that stores the previously returned values.
If we call the addFunction function again with the same argument, it first checks whether it has already gotten that value in its cache. If that's the case, the caches value will be returned, which saves on execution time. Else, if it's not cached, it will calculate the value and store it afterwards.
We call the addFunction function three times with the same value: on the first invocation, the value of the function when num is equal to 10 isn't cached yet. The condition of the if-statement num in cache returns false, and the else block gets executed: Calculated! 20 gets logged, and the value of the result gets added to the cache object. cache now looks like { 10: 20 }.
The second time, the cache object contains the value that gets returned for 10. The condition of the if-statement num in cache returns true, and 'From cache! 20' gets logged.
The third time, we pass 5 * 2 to the function which gets evaluated to 10. The cache object contains the value that gets returned for 10. The condition of the if-statement num in cache returns true, and 'From cache! 20' gets logged.
const myLifeSummedUp = ['โ', '๐ป', '๐ท', '๐ซ'];
for (let item in myLifeSummedUp) {
console.log(item);
}
for (let item of myLifeSummedUp) {
console.log(item);
}- A:
0123and"โ""๐ป""๐ท""๐ซ" - B:
"โ""๐ป""๐ท""๐ซ"and"โ""๐ป""๐ท""๐ซ" - C:
"โ""๐ป""๐ท""๐ซ"and0123 - D:
0123and{0: "โ", 1: "๐ป", 2: "๐ท", 3: "๐ซ"}
Answer
With a for-in loop, we can iterate over enumerable properties. In an array, the enumerable properties are the "keys" of array elements, which are actually their indexes. You could see an array as:
{0: "โ", 1: "๐ป", 2: "๐ท", 3: "๐ซ"}
Where the keys are the enumerable properties. 0 1 2 3 get logged.
With a for-of loop, we can iterate over iterables. An array is an iterable. When we iterate over the array, the variable "item" is equal to the element it's currently iterating over, "โ" "๐ป" "๐ท" "๐ซ" get logged.
const list = [1 + 2, 1 * 2, 1 / 2];
console.log(list);- A:
["1 + 2", "1 * 2", "1 / 2"] - B:
["12", 2, 0.5] - C:
[3, 2, 0.5] - D:
[1, 1, 1]
Answer
Array elements can hold any value. Numbers, strings, objects, other arrays, null, boolean values, undefined, and other expressions such as dates, functions, and calculations.
The element will be equal to the returned value. 1 + 2 returns 3, 1 * 2 returns 2, and 1 / 2 returns 0.5.
function sayHi(name) {
return `Hi there, ${name}`;
}
console.log(sayHi());- A:
Hi there, - B:
Hi there, undefined - C:
Hi there, null - D:
ReferenceError
Answer
By default, arguments have the value of undefined, unless a value has been passed to the function. In this case, we didn't pass a value for the name argument. name is equal to undefined which gets logged.
In ES6, we can overwrite this default undefined value with default parameters. For example:
function sayHi(name = "Lydia") { ... }
In this case, if we didn't pass a value or if we passed undefined, name would always be equal to the string Lydia
var status = '๐';
setTimeout(() => {
const status = '๐';
const data = {
status: '๐ฅ',
getStatus() {
return this.status;
},
};
console.log(data.getStatus());
console.log(data.getStatus.call(this));
}, 0);- A:
"๐ฅ"and"๐" - B:
"๐ฅ"and"๐" - C:
"๐"and"๐" - D:
"๐"and"๐"
Answer
The value of the this keyword is dependent on where you use it. In a method, like the getStatus method, the this keyword refers to the object that the method belongs to. The method belongs to the data object, so this refers to the data object. When we log this.status, the status property on the data object gets logged, which is "๐ฅ".
With the call method, we can change the object to which the this keyword refers. In functions, the this keyword refers to the the object that the function belongs to. We declared the setTimeout function on the global object, so within the setTimeout function, the this keyword refers to the global object. On the global object, there is a variable called status with the value of "๐". When logging this.status, "๐" gets logged.
const person = {
name: 'Lydia',
age: 21,
};
let city = person.city;
city = 'Amsterdam';
console.log(person);- A:
{ name: "Lydia", age: 21 } - B:
{ name: "Lydia", age: 21, city: "Amsterdam" } - C:
{ name: "Lydia", age: 21, city: undefined } - D:
"Amsterdam"
Answer
We set the variable city equal to the value of the property called city on the person object. There is no property on this object called city, so the variable city has the value of undefined.
Note that we are not referencing the person object itself! We simply set the variable city equal to the current value of the city property on the person object.
Then, we set city equal to the string "Amsterdam". This doesn't change the person object: there is no reference to that object.
When logging the person object, the unmodified object gets returned.
function checkAge(age) {
if (age < 18) {
const message = "Sorry, you're too young.";
} else {
const message = "Yay! You're old enough!";
}
return message;
}
console.log(checkAge(21));- A:
"Sorry, you're too young." - B:
"Yay! You're old enough!" - C:
ReferenceError - D:
undefined
Answer
Variables with the const and let keyword are block-scoped. A block is anything between curly brackets ({ }). In this case, the curly brackets of the if/else statements. You cannot reference a variable outside of the block it's declared in, a ReferenceError gets thrown.
fetch('https://www.website.com/api/user/1')
.then(res => res.json())
.then(res => console.log(res));- A: The result of the
fetchmethod. - B: The result of the second invocation of the
fetchmethod. - C: The result of the callback in the previous
.then(). - D: It would always be undefined.
Answer
The value of res in the second .then is equal to the returned value of the previous .then. You can keep chaining .thens like this, where the value is passed to the next handler.
86. Which option is a way to set hasName equal to true, provided you cannot pass true as an argument?
function getName(name) {
const hasName = //
}- A:
!!name - B:
name - C:
new Boolean(name) - D:
name.length
Answer
With !!name, we determine whether the value of name is truthy or falsy. If name is truthy, which we want to test for, !name returns false. !false (which is what !!name practically is) returns true.
By setting hasName equal to name, you set hasName equal to whatever value you passed to the getName function, not the boolean value true.
new Boolean(true) returns an object wrapper, not the boolean value itself.
name.length returns the length of the passed argument, not whether it's true.
console.log('I want pizza'[0]);- A:
""" - B:
"I" - C:
SyntaxError - D:
undefined
Answer
In order to get a character at a specific index of a string, you can use bracket notation. The first character in the string has index 0, and so on. In this case, we want to get the element with index 0, the character "I', which gets logged.
Note that this method is not supported in IE7 and below. In that case, use .charAt().
function sum(num1, num2 = num1) {
console.log(num1 + num2);
}
sum(10);- A:
NaN - B:
20 - C:
ReferenceError - D:
undefined
Answer
You can set a default parameter's value equal to another parameter of the function, as long as they've been defined before the default parameter. We pass the value 10 to the sum function. If the sum function only receives 1 argument, it means that the value for num2 is not passed, and the value of num1 is equal to the passed value 10 in this case. The default value of num2 is the value of num1, which is 10. num1 + num2 returns 20.
If you're trying to set a default parameter's value equal to a parameter which is defined after (to the right), the parameter's value hasn't been initialized yet, which will throw an error.
// module.js
export default () => 'Hello world';
export const name = 'Lydia';
// index.js
import * as data from './module';
console.log(data);- A:
{ default: function default(), name: "Lydia" } - B:
{ default: function default() } - C:
{ default: "Hello world", name: "Lydia" } - D: Global object of
module.js
Answer
With the import * as name syntax, we import all exports from the module.js file into the index.js file as a new object called data is created. In the module.js file, there are two exports: the default export, and a named export. The default export is a function which returns the string "Hello World", and the named export is a variable called name which has the value of the string "Lydia".
The data object has a default property for the default export, other properties have the names of the named exports and their corresponding values.
class Person {
constructor(name) {
this.name = name;
}
}
const member = new Person('John');
console.log(typeof member);- A:
"class" - B:
"function" - C:
"object" - D:
"string"
Answer
Classes are syntactical sugar for function constructors. The equivalent of the Person class as a function constructor would be:
function Person(name) {
this.name = name;
}Calling a function constructor with new results in the creation of an instance of Person, typeof keyword returns "object" for an instance. typeof member returns "object".
let newList = [1, 2, 3].push(4);
console.log(newList.push(5));- A:
[1, 2, 3, 4, 5] - B:
[1, 2, 3, 5] - C:
[1, 2, 3, 4] - D:
Error
Answer
The .push method returns the new length of the array, not the array itself! By setting newList equal to [1, 2, 3].push(4), we set newList equal to the new length of the array: 4.
Then, we try to use the .push method on newList. Since newList is the numerical value 4, we cannot use the .push method: a TypeError is thrown.
function giveLydiaPizza() {
return 'Here is pizza!';
}
const giveLydiaChocolate = () =>
"Here's chocolate... now go hit the gym already.";
console.log(giveLydiaPizza.prototype);
console.log(giveLydiaChocolate.prototype);- A:
{ constructor: ...}{ constructor: ...} - B:
{}{ constructor: ...} - C:
{ constructor: ...}{} - D:
{ constructor: ...}undefined
Answer
Regular functions, such as the giveLydiaPizza function, have a prototype property, which is an object (prototype object) with a constructor property. Arrow functions however, such as the giveLydiaChocolate function, do not have this prototype property. undefined gets returned when trying to access the prototype property using giveLydiaChocolate.prototype.
const person = {
name: 'Lydia',
age: 21,
};
for (const [x, y] of Object.entries(person)) {
console.log(x, y);
}- A:
nameLydiaandage21 - B:
["name", "Lydia"]and["age", 21] - C:
["name", "age"]andundefined - D:
Error
Answer
Object.entries(person) returns an array of nested arrays, containing the keys and objects:
[ [ 'name', 'Lydia' ], [ 'age', 21 ] ]
Using the for-of loop, we can iterate over each element in the array, the subarrays in this case. We can destructure the subarrays instantly in the for-of loop, using const [x, y]. x is equal to the first element in the subarray, y is equal to the second element in the subarray.
The first subarray is [ "name", "Lydia" ], with x equal to "name", and y equal to "Lydia", which get logged.
The second subarray is [ "age", 21 ], with x equal to "age", and y equal to 21, which get logged.
function getItems(fruitList, ...args, favoriteFruit) {
return [...fruitList, ...args, favoriteFruit]
}
getItems(["banana", "apple"], "pear", "orange")- A:
["banana", "apple", "pear", "orange"] - B:
[["banana", "apple"], "pear", "orange"] - C:
["banana", "apple", ["pear"], "orange"] - D:
SyntaxError
Answer
...args is a rest parameter. The rest parameter's value is an array containing all remaining arguments, and can only be the last parameter! In this example, the rest parameter was the second parameter. This is not possible, and will throw a syntax error.
function getItems(fruitList, favoriteFruit, ...args) {
return [...fruitList, ...args, favoriteFruit];
}
getItems(['banana', 'apple'], 'pear', 'orange');The above example works. This returns the array [ 'banana', 'apple', 'orange', 'pear' ]
function nums(a, b) {
if (a > b) console.log('a is bigger');
else console.log('b is bigger');
return
a + b;
}
console.log(nums(4, 2));
console.log(nums(1, 2));- A:
a is bigger,6andb is bigger,3 - B:
a is bigger,undefinedandb is bigger,undefined - C:
undefinedandundefined - D:
SyntaxError
Answer
In JavaScript, we don't have to write the semicolon (;) explicitly, however the JavaScript engine still adds them after statements. This is called Automatic Semicolon Insertion. A statement can for example be variables, or keywords like throw, return, break, etc.
Here, we wrote a return statement, and another value a + b on a new line. However, since it's a new line, the engine doesn't know that it's actually the value that we wanted to return. Instead, it automatically added a semicolon after return. You could see this as:
return;
a + b;This means that a + b is never reached, since a function stops running after the return keyword. If no value gets returned, like here, the function returns undefined. Note that there is no automatic insertion after if/else statements!
class Person {
constructor() {
this.name = 'Lydia';
}
}
Person = class AnotherPerson {
constructor() {
this.name = 'Sarah';
}
};
const member = new Person();
console.log(member.name);- A:
"Lydia" - B:
"Sarah" - C:
Error: cannot redeclare Person - D:
SyntaxError
Answer
We can set classes equal to other classes/function constructors. In this case, we set Person equal to AnotherPerson. The name on this constructor is Sarah, so the name property on the new Person instance member is "Sarah".
const info = {
[Symbol('a')]: 'b',
};
console.log(info);
console.log(Object.keys(info));- A:
{Symbol('a'): 'b'}and["{Symbol('a')"] - B:
{}and[] - C:
{ a: "b" }and["a"] - D:
{Symbol('a'): 'b'}and[]
Answer
A Symbol is not enumerable. The Object.keys method returns all enumerable key properties on an object. The Symbol won't be visible, and an empty array is returned. When logging the entire object, all properties will be visible, even non-enumerable ones.
This is one of the many qualities of a symbol: besides representing an entirely unique value (which prevents accidental name collision on objects, for example when working with 2 libraries that want to add properties to the same object), you can also "hide" properties on objects this way (although not entirely. You can still access symbols using the Object.getOwnPropertySymbols() method).
const getList = ([x, ...y]) => [x, y]
const getUser = user => { name: user.name, age: user.age }
const list = [1, 2, 3, 4]
const user = { name: "Lydia", age: 21 }
console.log(getList(list))
console.log(getUser(user))- A:
[1, [2, 3, 4]]andSyntaxError - B:
[1, [2, 3, 4]]and{ name: "Lydia", age: 21 } - C:
[1, 2, 3, 4]and{ name: "Lydia", age: 21 } - D:
Errorand{ name: "Lydia", age: 21 }
Answer
The getList function receives an array as its argument. Between the parentheses of the getList function, we destructure this array right away. You could see this as:
[x, ...y] = [1, 2, 3, 4]
With the rest parameter ...y, we put all "remaining" arguments in an array. The remaining arguments are 2, 3 and 4 in this case. The value of y is an array, containing all the rest parameters. The value of x is equal to 1 in this case, so when we log [x, y], [1, [2, 3, 4]] gets logged.
The getUser function receives an object. With arrow functions, we don't have to write curly brackets if we just return one value. However, if you want to instantly return an object from an arrow function, you have to write it between parentheses, otherwise everything between the two braces will be interpreted as a block statement. In this case the code between the braces is not a valid JavaScript code, so a SyntaxError gets thrown.
The following function would have returned an object:
const getUser = user => ({ name: user.name, age: user.age })
const name = 'Lydia';
console.log(name());- A:
SyntaxError - B:
ReferenceError - C:
TypeError - D:
undefined
Answer
The variable name holds the value of a string, which is not a function, thus cannot invoke.
TypeErrors get thrown when a value is not of the expected type. JavaScript expected name to be a function since we're trying to invoke it. It was a string however, so a TypeError gets thrown: name is not a function!
SyntaxErrors get thrown when you've written something that isn't valid JavaScript, for example when you've written the word return as retrun.
ReferenceErrors get thrown when JavaScript isn't able to find a reference to a value that you're trying to access.
// ๐โจ This is my 100th question! โจ๐
const output = `${[] && 'Im'}possible!
You should${'' && `n't`} see a therapist after so much JavaScript lol`;- A:
possible! You should see a therapist after so much JavaScript lol - B:
Impossible! You should see a therapist after so much JavaScript lol - C:
possible! You shouldn't see a therapist after so much JavaScript lol - D:
Impossible! You shouldn't see a therapist after so much JavaScript lol
Answer
[] is a truthy value. With the && operator, the right-hand value will be returned if the left-hand value is a truthy value. In this case, the left-hand value [] is a truthy value, so "Im' gets returned.
"" is a falsy value. If the left-hand value is falsy, nothing gets returned. n't doesn't get returned.
const one = false || {} || null;
const two = null || false || '';
const three = [] || 0 || true;
console.log(one, two, three);- A:
falsenull[] - B:
null""true - C:
{}""[] - D:
nullnulltrue
Answer
With the || operator, we can return the first truthy operand. If all values are falsy, the last operand gets returned.
(false || {} || null): the empty object {} is a truthy value. This is the first (and only) truthy value, which gets returned. one is equal to {}.
(null || false || ""): all operands are falsy values. This means that the last operand, "" gets returned. two is equal to "".
([] || 0 || ""): the empty array[] is a truthy value. This is the first truthy value, which gets returned. three is equal to [].
const myPromise = () => Promise.resolve('I have resolved!');
function firstFunction() {
myPromise().then(res => console.log(res));
console.log('second');
}
async function secondFunction() {
console.log(await myPromise());
console.log('second');
}
firstFunction();
secondFunction();- A:
I have resolved!,secondandI have resolved!,second - B:
second,I have resolved!andsecond,I have resolved! - C:
I have resolved!,secondandsecond,I have resolved! - D:
second,I have resolved!andI have resolved!,second
Answer
With a promise, we basically say I want to execute this function, but I'll put it aside for now while it's running since this might take a while. Only when a certain value is resolved (or rejected), and when the call stack is empty, I want to use this value.
We can get this value with both .then and the await keyword in an async function. Although we can get a promise's value with both .then and await, they work a bit differently.
In the firstFunction, we (sort of) put the myPromise function aside while it was running, but continued running the other code, which is console.log('second') in this case. Then, the function resolved with the string I have resolved, which then got logged after it saw that the callstack was empty.
With the await keyword in secondFunction, we literally pause the execution of an async function until the value has been resolved before moving to the next line.
This means that it waited for the myPromise to resolve with the value I have resolved, and only once that happened, we moved to the next line: second got logged.
const set = new Set();
set.add(1);
set.add('Lydia');
set.add({ name: 'Lydia' });
for (let item of set) {
console.log(item + 2);
}- A:
3,NaN,NaN - B:
3,7,NaN - C:
3,Lydia2,[object Object]2 - D:
"12",Lydia2,[object Object]2
Answer
The + operator is not only used for adding numerical values, but we can also use it to concatenate strings. Whenever the JavaScript engine sees that one or more values are not a number, it coerces the number into a string.
The first one is 1, which is a numerical value. 1 + 2 returns the number 3.
However, the second one is a string "Lydia". "Lydia" is a string and 2 is a number: 2 gets coerced into a string. "Lydia" and "2" get concatenated, which results in the string "Lydia2".
{ name: "Lydia" } is an object. Neither a number nor an object is a string, so it stringifies both. Whenever we stringify a regular object, it becomes "[object Object]". "[object Object]" concatenated with "2" becomes "[object Object]2".
Promise.resolve(5);- A:
5 - B:
Promise {<pending>: 5} - C:
Promise {<fulfilled>: 5} - D:
Error
Answer
We can pass any type of value we want to Promise.resolve, either a promise or a non-promise. The method itself returns a promise with the resolved value (<fulfilled>). If you pass a regular function, it'll be a resolved promise with a regular value. If you pass a promise, it'll be a resolved promise with the resolved value of that passed promise.
In this case, we just passed the numerical value 5. It returns a resolved promise with the value 5.
function compareMembers(person1, person2 = person) {
if (person1 !== person2) {
console.log('Not the same!');
} else {
console.log('They are the same!');
}
}
const person = { name: 'Lydia' };
compareMembers(person);- A:
Not the same! - B:
They are the same! - C:
ReferenceError - D:
SyntaxError
Answer
Objects are passed by reference. When we check objects for strict equality (===), we're comparing their references.
We set the default value for person2 equal to the person object, and passed the person object as the value for person1.
This means that both values have a reference to the same spot in memory, thus they are equal.
The code block in the else statement gets run, and They are the same! gets logged.
const colorConfig = {
red: true,
blue: false,
green: true,
black: true,
yellow: false,
};
const colors = ['pink', 'red', 'blue'];
console.log(colorConfig.colors[1]);- A:
true - B:
false - C:
undefined - D:
TypeError
Answer
In JavaScript, we have two ways to access properties on an object: bracket notation, or dot notation. In this example, we use dot notation (colorConfig.colors) instead of bracket notation (colorConfig["colors"]).
With dot notation, JavaScript tries to find the property on the object with that exact name. In this example, JavaScript tries to find a property called colors on the colorConfig object. There is no property called colors, so this returns undefined. Then, we try to access the value of the first element by using [1]. We cannot do this on a value that's undefined, so it throws a TypeError: Cannot read property '1' of undefined.
JavaScript interprets (or unboxes) statements. When we use bracket notation, it sees the first opening bracket [ and keeps going until it finds the closing bracket ]. Only then, it will evaluate the statement. If we would've used colorConfig[colors[1]], it would have returned the value of the red property on the colorConfig object.
console.log('โค๏ธ' === 'โค๏ธ');- A:
true - B:
false
Answer
Under the hood, emojis are unicodes. The unicodes for the heart emoji is "U+2764 U+FE0F". These are always the same for the same emojis, so we're comparing two equal strings to each other, which returns true.
const emojis = ['โจ', '๐ฅ', '๐'];
emojis.map(x => x + 'โจ');
emojis.filter(x => x !== '๐ฅ');
emojis.find(x => x !== '๐ฅ');
emojis.reduce((acc, cur) => acc + 'โจ');
emojis.slice(1, 2, 'โจ');
emojis.splice(1, 2, 'โจ');- A:
All of them - B:
mapreduceslicesplice - C:
mapslicesplice - D:
splice
Answer
With splice method, we modify the original array by deleting, replacing or adding elements. In this case, we removed 2 items from index 1 (we removed '๐ฅ' and '๐') and added the โจ emoji instead.
map, filter and slice return a new array, find returns an element, and reduce returns a reduced value.
const food = ['๐', '๐ซ', '๐ฅ', '๐'];
const info = { favoriteFood: food[0] };
info.favoriteFood = '๐';
console.log(food);- A:
['๐', '๐ซ', '๐ฅ', '๐'] - B:
['๐', '๐ซ', '๐ฅ', '๐'] - C:
['๐', '๐', '๐ซ', '๐ฅ', '๐'] - D:
ReferenceError
Answer
We set the value of the favoriteFood property on the info object equal to the string with the pizza emoji, '๐'. A string is a primitive data type. In JavaScript, primitive data types don't interact by reference.
In JavaScript, primitive data types (everything that's not an object) interact by value. In this case, we set the value of the favoriteFood property on the info object equal to the value of the first element in the food array, the string with the pizza emoji in this case ('๐'). A string is a primitive data type, and interact by value (see my blogpost if you're interested in learning more)
Then, we change the value of the favoriteFood property on the info object. The food array hasn't changed, since the value of favoriteFood was merely a copy of the value of the first element in the array, and doesn't have a reference to the same spot in memory as the element on food[0]. When we log food, it's still the original array, ['๐', '๐ซ', '๐ฅ', '๐'].
JSON.parse();- A: Parses JSON to a JavaScript value
- B: Parses a JavaScript object to JSON
- C: Parses any JavaScript value to JSON
- D: Parses JSON to a JavaScript object only
Answer
With the JSON.parse() method, we can parse JSON string to a JavaScript value.
// Stringifying a number into valid JSON, then parsing the JSON string to a JavaScript value:
const jsonNumber = JSON.stringify(4); // '4'
JSON.parse(jsonNumber); // 4
// Stringifying an array value into valid JSON, then parsing the JSON string to a JavaScript value:
const jsonArray = JSON.stringify([1, 2, 3]); // '[1, 2, 3]'
JSON.parse(jsonArray); // [1, 2, 3]
// Stringifying an object into valid JSON, then parsing the JSON string to a JavaScript value:
const jsonArray = JSON.stringify({ name: 'Lydia' }); // '{"name":"Lydia"}'
JSON.parse(jsonArray); // { name: 'Lydia' }let name = 'Lydia';
function getName() {
console.log(name);
let name = 'Sarah';
}
getName();- A: Lydia
- B: Sarah
- C:
undefined - D:
ReferenceError
Answer
Each function has its own execution context (or scope). The getName function first looks within its own context (scope) to see if it contains the variable name we're trying to access. In this case, the getName function contains its own name variable: we declare the variable name with the let keyword, and with the value of 'Sarah'.
Variables with the let keyword (and const) are hoisted, but unlike var, don't get initialized. They are not accessible before the line we declare (initialize) them. This is called the "temporal dead zone". When we try to access the variables before they are declared, JavaScript throws a ReferenceError.
If we wouldn't have declared the name variable within the getName function, the javascript engine would've looked down the scope chain. The outer scope has a variable called name with the value of Lydia. In that case, it would've logged Lydia.
let name = 'Lydia';
function getName() {
console.log(name);
}
getName(); // Lydiafunction* generatorOne() {
yield ['a', 'b', 'c'];
}
function* generatorTwo() {
yield* ['a', 'b', 'c'];
}
const one = generatorOne();
const two = generatorTwo();
console.log(one.next().value);
console.log(two.next().value);- A:
aanda - B:
aandundefined - C:
['a', 'b', 'c']anda - D:
aand['a', 'b', 'c']
Answer
With the yield keyword, we yield values in a generator function. With the yield* keyword, we can yield values from another generator function, or iterable object (for example an array).
In generatorOne, we yield the entire array ['a', 'b', 'c'] using the yield keyword. The value of value property on the object returned by the next method on one (one.next().value) is equal to the entire array ['a', 'b', 'c'].
console.log(one.next().value); // ['a', 'b', 'c']
console.log(one.next().value); // undefinedIn generatorTwo, we use the yield* keyword. This means that the first yielded value of two, is equal to the first yielded value in the iterator. The iterator is the array ['a', 'b', 'c']. The first yielded value is a, so the first time we call two.next().value, a is returned.
console.log(two.next().value); // 'a'
console.log(two.next().value); // 'b'
console.log(two.next().value); // 'c'
console.log(two.next().value); // undefinedconsole.log(`${(x => x)('I love')} to program`);- A:
I love to program - B:
undefined to program - C:
${(x => x)('I love') to program - D:
TypeError
Answer
Expressions within template literals are evaluated first. This means that the string will contain the returned value of the expression, the immediately invoked function (x => x)('I love') in this case. We pass the value 'I love' as an argument to the x => x arrow function. x is equal to 'I love', which gets returned. This results in I love to program.
let config = {
alert: setInterval(() => {
console.log('Alert!');
}, 1000),
};
config = null;- A: The
setIntervalcallback won't be invoked - B: The
setIntervalcallback gets invoked once - C: The
setIntervalcallback will still be called every second - D: We never invoked
config.alert(), config isnull
Answer
Normally when we set objects equal to null, those objects get garbage collected as there is no reference anymore to that object. However, since the callback function within setInterval is an arrow function (thus bound to the config object), the callback function still holds a reference to the config object.
As long as there is a reference, the object won't get garbage collected.
Since this is an interval, setting config to null or delete-ing config.alert won't garbage-collect the interval, so the interval will still be called.
It should be cleared with clearInterval(config.alert) to remove it from memory.
Since it was not cleared, the setInterval callback function will still get invoked every 1000ms (1s).
const myMap = new Map();
const myFunc = () => 'greeting';
myMap.set(myFunc, 'Hello world!');
//1
myMap.get('greeting');
//2
myMap.get(myFunc);
//3
myMap.get(() => 'greeting');- A: 1
- B: 2
- C: 2 and 3
- D: All of them
Answer
When adding a key/value pair using the set method, the key will be the value of the first argument passed to the set function, and the value will be the second argument passed to the set function. The key is the function () => 'greeting' in this case, and the value 'Hello world'. myMap is now { () => 'greeting' => 'Hello world!' }.
1 is wrong, since the key is not 'greeting' but () => 'greeting'.
3 is wrong, since we're creating a new function by passing it as a parameter to the get method. Object interact by reference. Functions are objects, which is why two functions are never strictly equal, even if they are identical: they have a reference to a different spot in memory.
const person = {
name: 'Lydia',
age: 21,
};
const changeAge = (x = { ...person }) => (x.age += 1);
const changeAgeAndName = (x = { ...person }) => {
x.age += 1;
x.name = 'Sarah';
};
changeAge(person);
changeAgeAndName();
console.log(person);- A:
{name: "Sarah", age: 22} - B:
{name: "Sarah", age: 23} - C:
{name: "Lydia", age: 22} - D:
{name: "Lydia", age: 23}
Answer
Both the changeAge and changeAgeAndName functions have a default parameter, namely a newly created object { ...person }. This object has copies of all the key/values in the person object.
First, we invoke the changeAge function and pass the person object as its argument. This function increases the value of the age property by 1. person is now { name: "Lydia", age: 22 }.
Then, we invoke the changeAgeAndName function, however we don't pass a parameter. Instead, the value of x is equal to a new object: { ...person }. Since it's a new object, it doesn't affect the values of the properties on the person object. person is still equal to { name: "Lydia", age: 22 }.
function sumValues(x, y, z) {
return x + y + z;
}- A:
sumValues([...1, 2, 3]) - B:
sumValues([...[1, 2, 3]]) - C:
sumValues(...[1, 2, 3]) - D:
sumValues([1, 2, 3])
Answer
With the spread operator ..., we can spread iterables to individual elements. The sumValues function receives three arguments: x, y and z. ...[1, 2, 3] will result in 1, 2, 3, which we pass to the sumValues function.
let num = 1;
const list = ['๐ฅณ', '๐ค ', '๐ฅฐ', '๐คช'];
console.log(list[(num += 1)]);- A:
๐ค - B:
๐ฅฐ - C:
SyntaxError - D:
ReferenceError
Answer
With the += operand, we're incrementing the value of num by 1. num had the initial value 1, so 1 + 1 is 2. The item on the second index in the list array is ๐ฅฐ, console.log(list[2]) prints ๐ฅฐ.
const person = {
firstName: 'Lydia',
lastName: 'Hallie',
pet: {
name: 'Mara',
breed: 'Dutch Tulip Hound',
},
getFullName() {
return `${this.firstName} ${this.lastName}`;
},
};
console.log(person.pet?.name);
console.log(person.pet?.family?.name);
console.log(person.getFullName?.());
console.log(member.getLastName?.());- A:
undefinedundefinedundefinedundefined - B:
MaraundefinedLydia HallieReferenceError - C:
MaranullLydia Hallienull - D:
nullReferenceErrornullReferenceError
Answer
With the optional chaining operator ?., we no longer have to explicitly check whether the deeper nested values are valid or not. If we're trying to access a property on an undefined or null value (nullish), the expression short-circuits and returns undefined.
person.pet?.name: person has a property named pet: person.pet is not nullish. It has a property called name, and returns Mara.
person.pet?.family?.name: person has a property named pet: person.pet is not nullish. pet does not have a property called family, person.pet.family is nullish. The expression returns undefined.
person.getFullName?.(): person has a property named getFullName: person.getFullName() is not nullish and can get invoked, which returns Lydia Hallie.
member.getLastName?.(): variable member is non existent therefore a ReferenceError gets thrown!
const groceries = ['banana', 'apple', 'peanuts'];
if (groceries.indexOf('banana')) {
console.log('We have to buy bananas!');
} else {
console.log(`We don't have to buy bananas!`);
}- A: We have to buy bananas!
- B: We don't have to buy bananas
- C:
undefined - D:
1
Answer
We passed the condition groceries.indexOf("banana") to the if-statement. groceries.indexOf("banana") returns 0, which is a falsy value. Since the condition in the if-statement is falsy, the code in the else block runs, and We don't have to buy bananas! gets logged.
const config = {
languages: [],
set language(lang) {
return this.languages.push(lang);
},
};
console.log(config.language);- A:
function language(lang) { this.languages.push(lang } - B:
0 - C:
[] - D:
undefined
Answer
The language method is a setter. Setters don't hold an actual value, their purpose is to modify properties. When calling a setter method, undefined gets returned.
const name = 'Lydia Hallie';
console.log(!typeof name === 'object');
console.log(!typeof name === 'string');- A:
falsetrue - B:
truefalse - C:
falsefalse - D:
truetrue
Answer
typeof name returns "string". The string "string" is a truthy value, so !typeof name returns the boolean value false. false === "object" and false === "string" both returnfalse.
(If we wanted to check whether the type was (un)equal to a certain type, we should've written !== instead of !typeof)
const add = x => y => z => {
console.log(x, y, z);
return x + y + z;
};
add(4)(5)(6);- A:
456 - B:
654 - C:
4functionfunction - D:
undefinedundefined6
Answer
The add function returns an arrow function, which returns an arrow function, which returns an arrow function (still with me?). The first function receives an argument x with the value of 4. We invoke the second function, which receives an argument y with the value 5. Then we invoke the third function, which receives an argument z with the value 6. When we're trying to access the value x, y and z within the last arrow function, the JS engine goes up the scope chain in order to find the values for x and y accordingly. This returns 4 5 6.
async function* range(start, end) {
for (let i = start; i <= end; i++) {
yield Promise.resolve(i);
}
}
(async () => {
const gen = range(1, 3);
for await (const item of gen) {
console.log(item);
}
})();- A:
Promise {1}Promise {2}Promise {3} - B:
Promise {<pending>}Promise {<pending>}Promise {<pending>} - C:
123 - D:
undefinedundefinedundefined
Answer
The generator function range returns an async object with promises for each item in the range we pass: Promise{1}, Promise{2}, Promise{3}. We set the variable gen equal to the async object, after which we loop over it using a for await ... of loop. We set the variable item equal to the returned Promise values: first Promise{1}, then Promise{2}, then Promise{3}. Since we're awaiting the value of item, the resolved promise, the resolved values of the promises get returned: 1, 2, then 3.
const myFunc = ({ x, y, z }) => {
console.log(x, y, z);
};
myFunc(1, 2, 3);- A:
123 - B:
{1: 1}{2: 2}{3: 3} - C:
{ 1: undefined }undefinedundefined - D:
undefinedundefinedundefined
Answer
myFunc expects an object with properties x, y and z as its argument. Since we're only passing three separate numeric values (1, 2, 3) instead of one object with properties x, y and z ({x: 1, y: 2, z: 3}), x, y and z have their default value of undefined.
function getFine(speed, amount) {
const formattedSpeed = new Intl.NumberFormat('en-US', {
style: 'unit',
unit: 'mile-per-hour'
}).format(speed);
const formattedAmount = new Intl.NumberFormat('en-US', {
style: 'currency',
currency: 'USD'
}).format(amount);
return `The driver drove ${formattedSpeed} and has to pay ${formattedAmount}`;
}
console.log(getFine(130, 300))- A: The driver drove 130 and has to pay 300
- B: The driver drove 130 mph and has to pay $300.00
- C: The driver drove undefined and has to pay undefined
- D: The driver drove 130.00 and has to pay 300.00
Answer
With the Intl.NumberFormat method, we can format numeric values to any locale. We format the numeric value 130 to the en-US locale as a unit in mile-per-hour, which results in 130 mph. The numeric value 300 to the en-US locale as a currency in USD results in $300.00.
const spookyItems = ['๐ป', '๐', '๐ธ'];
({ item: spookyItems[3] } = { item: '๐' });
console.log(spookyItems);- A:
["๐ป", "๐", "๐ธ"] - B:
["๐ป", "๐", "๐ธ", "๐"] - C:
["๐ป", "๐", "๐ธ", { item: "๐" }] - D:
["๐ป", "๐", "๐ธ", "[object Object]"]
Answer
By destructuring objects, we can unpack values from the right-hand object, and assign the unpacked value to the value of the same property name on the left-hand object. In this case, we're assigning the value "๐" to spookyItems[3]. This means that we're modifying the spookyItems array, we're adding the "๐" to it. When logging spookyItems, ["๐ป", "๐", "๐ธ", "๐"] gets logged.
const name = 'Lydia Hallie';
const age = 21;
console.log(Number.isNaN(name));
console.log(Number.isNaN(age));
console.log(isNaN(name));
console.log(isNaN(age));- A:
truefalsetruefalse - B:
truefalsefalsefalse - C:
falsefalsetruefalse - D:
falsetruefalsetrue
Answer
With the Number.isNaN method, you can check if the value you pass is a numeric value and equal to NaN. name is not a numeric value, so Number.isNaN(name) returns false. age is a numeric value, but is not equal to NaN, so Number.isNaN(age) returns false.
With the isNaN method, you can check if the value you pass is not a number. name is not a number, so isNaN(name) returns true. age is a number, so isNaN(age) returns false.
const randomValue = 21;
function getInfo() {
console.log(typeof randomValue);
const randomValue = 'Lydia Hallie';
}
getInfo();- A:
"number" - B:
"string" - C:
undefined - D:
ReferenceError
Answer
Variables declared with the const keyword are not referenceable before their initialization: this is called the temporal dead zone. In the getInfo function, the variable randomValue is scoped in the functional scope of getInfo. On the line where we want to log the value of typeof randomValue, the variable randomValue isn't initialized yet: a ReferenceError gets thrown! The engine didn't go down the scope chain since we declared the variable randomValue in the getInfo function.
const myPromise = Promise.resolve('Woah some cool data');
(async () => {
try {
console.log(await myPromise);
} catch {
throw new Error(`Oops didn't work`);
} finally {
console.log('Oh finally!');
}
})();- A:
Woah some cool data - B:
Oh finally! - C:
Woah some cool dataOh finally! - D:
Oops didn't workOh finally!
Answer
In the try block, we're logging the awaited value of the myPromise variable: "Woah some cool data". Since no errors were thrown in the try block, the code in the catch block doesn't run. The code in the finally block always runs, "Oh finally!" gets logged.
const emojis = ['๐ฅ', ['โจ', 'โจ', ['๐', '๐']]];
console.log(emojis.flat(1));- A:
['๐ฅ', ['โจ', 'โจ', ['๐', '๐']]] - B:
['๐ฅ', 'โจ', 'โจ', ['๐', '๐']] - C:
['๐ฅ', ['โจ', 'โจ', '๐', '๐']] - D:
['๐ฅ', 'โจ', 'โจ', '๐', '๐']
Answer
With the flat method, we can create a new, flattened array. The depth of the flattened array depends on the value that we pass. In this case, we passed the value 1 (which we didn't have to, that's the default value), meaning that only the arrays on the first depth will be concatenated. ['๐ฅ'] and ['โจ', 'โจ', ['๐', '๐']] in this case. Concatenating these two arrays results in ['๐ฅ', 'โจ', 'โจ', ['๐', '๐']].
class Counter {
constructor() {
this.count = 0;
}
increment() {
this.count++;
}
}
const counterOne = new Counter();
counterOne.increment();
counterOne.increment();
const counterTwo = counterOne;
counterTwo.increment();
console.log(counterOne.count);- A:
0 - B:
1 - C:
2 - D:
3
Answer
counterOne is an instance of the Counter class. The counter class contains a count property on its constructor, and an increment method. First, we invoked the increment method twice by calling counterOne.increment(). Currently, counterOne.count is 2.
Then, we create a new variable counterTwo, and set it equal to counterOne. Since objects interact by reference, we're just creating a new reference to the same spot in memory that counterOne points to. Since it has the same spot in memory, any changes made to the object that counterTwo has a reference to, also apply to counterOne. Currently, counterTwo.count is 2.
We invoke counterTwo.increment(), which sets count to 3. Then, we log the count on counterOne, which logs 3.
const myPromise = Promise.resolve(Promise.resolve('Promise'));
function funcOne() {
setTimeout(() => console.log('Timeout 1!'), 0);
myPromise.then(res => res).then(res => console.log(`${res} 1!`));
console.log('Last line 1!');
}
async function funcTwo() {
const res = await myPromise;
console.log(`${res} 2!`)
setTimeout(() => console.log('Timeout 2!'), 0);
console.log('Last line 2!');
}
funcOne();
funcTwo();- A:
Promise 1! Last line 1! Promise 2! Last line 2! Timeout 1! Timeout 2! - B:
Last line 1! Timeout 1! Promise 1! Last line 2! Promise2! Timeout 2! - C:
Last line 1! Promise 2! Last line 2! Promise 1! Timeout 1! Timeout 2! - D:
Timeout 1! Promise 1! Last line 1! Promise 2! Timeout 2! Last line 2!
Answer
First, we invoke funcOne. On the first line of funcOne, we call the asynchronous setTimeout function, from which the callback is sent to the Web API. (see my article on the event loop here.)
Then we call the myPromise promise, which is an asynchronous operation.
Both the promise and the timeout are asynchronous operations, the function keeps on running while it's busy completing the promise and handling the setTimeout callback. This means that Last line 1! gets logged first, since this is not an asynchonous operation.
Since the callstack is not empty yet, the setTimeout function and promise in funcOne cannot get added to the callstack yet.
In funcTwo, the variable res gets Promise because Promise.resolve(Promise.resolve('Promise')) is equivalent to Promise.resolve('Promise') since resolving a promise just resolves it's value. The await in this line stops the execution of the function until it receives the resolution of the promise and then keeps on running synchronously until completion, so Promise 2! and then Last line 2! are logged and the setTimeout is sent to the Web API.
Then the call stack is empty. Promises are microtasks so they are resolved first when the call stack is empty so Promise 1! gets to be logged.
Now, since funcTwo popped off the call stack, the call stack is empty. The callbacks waiting in the queue (() => console.log("Timeout 1!") from funcOne, and () => console.log("Timeout 2!") from funcTwo) get added to the call stack one by one. The first callback logs Timeout 1!, and gets popped off the stack. Then, the second callback logs Timeout 2!, and gets popped off the stack.
// sum.js
export default function sum(x) {
return x + x;
}
// index.js
import * as sum from './sum';- A:
sum(4) - B:
sum.sum(4) - C:
sum.default(4) - D: Default aren't imported with
*, only named exports
Answer
With the asterisk *, we import all exported values from that file, both default and named. If we had the following file:
// info.js
export const name = 'Lydia';
export const age = 21;
export default 'I love JavaScript';
// index.js
import * as info from './info';
console.log(info);The following would get logged:
{
default: "I love JavaScript",
name: "Lydia",
age: 21
}For the sum example, it means that the imported value sum looks like this:
{ default: function sum(x) { return x + x } }We can invoke this function, by calling sum.default
const handler = {
set: () => console.log('Added a new property!'),
get: () => console.log('Accessed a property!'),
};
const person = new Proxy({}, handler);
person.name = 'Lydia';
person.name;- A:
Added a new property! - B:
Accessed a property! - C:
Added a new property!Accessed a property! - D: Nothing gets logged
Answer
With a Proxy object, we can add custom behavior to an object that we pass to it as the second argument. In this case, we pass the handler object which contained two properties: set and get. set gets invoked whenever we set property values, get gets invoked whenever we get (access) property values.
The first argument is an empty object {}, which is the value of person. To this object, the custom behavior specified in the handler object gets added. If we add a property to the person object, set will get invoked. If we access a property on the person object, get gets invoked.
First, we added a new property name to the proxy object (person.name = "Lydia"). set gets invoked, and logs "Added a new property!".
Then, we access a property value on the proxy object, the get property on the handler object got invoked. "Accessed a property!" gets logged.
const person = { name: 'Lydia Hallie' };
Object.seal(person);- A:
person.name = "Evan Bacon" - B:
person.age = 21 - C:
delete person.name - D:
Object.assign(person, { age: 21 })
Answer
With Object.seal we can prevent new properties from being added, or existing properties to be removed.
However, you can still modify the value of existing properties.
const person = {
name: 'Lydia Hallie',
address: {
street: '100 Main St',
},
};
Object.freeze(person);- A:
person.name = "Evan Bacon" - B:
delete person.address - C:
person.address.street = "101 Main St" - D:
person.pet = { name: "Mara" }
Answer
The Object.freeze method freezes an object. No properties can be added, modified, or removed.
However, it only shallowly freezes the object, meaning that only direct properties on the object are frozen. If the property is another object, like address in this case, the properties on that object aren't frozen, and can be modified.
const add = x => x + x;
function myFunc(num = 2, value = add(num)) {
console.log(num, value);
}
myFunc();
myFunc(3);- A:
24and36 - B:
2NaNand3NaN - C:
2Errorand36 - D:
24and3Error
Answer
First, we invoked myFunc() without passing any arguments. Since we didn't pass arguments, num and value got their default values: num is 2, and value the returned value of the function add. To the add function, we pass num as an argument, which had the value of 2. add returns 4, which is the value of value.
Then, we invoked myFunc(3) and passed the value 3 as the value for the argument num. We didn't pass an argument for value. Since we didn't pass a value for the value argument, it got the default value: the returned value of the add function. To add, we pass num, which has the value of 3. add returns 6, which is the value of value.
class Counter {
#number = 10
increment() {
this.#number++
}
getNum() {
return this.#number
}
}
const counter = new Counter()
counter.increment()
console.log(counter.#number)- A:
10 - B:
11 - C:
undefined - D:
SyntaxError
Answer
In ES2020, we can add private variables in classes by using the #. We cannot access these variables outside of the class. When we try to log counter.#number, a SyntaxError gets thrown: we cannot acccess it outside the Counter class!
const teams = [
{ name: 'Team 1', members: ['Paul', 'Lisa'] },
{ name: 'Team 2', members: ['Laura', 'Tim'] },
];
function* getMembers(members) {
for (let i = 0; i < members.length; i++) {
yield members[i];
}
}
function* getTeams(teams) {
for (let i = 0; i < teams.length; i++) {
// โจ SOMETHING IS MISSING HERE โจ
}
}
const obj = getTeams(teams);
obj.next(); // { value: "Paul", done: false }
obj.next(); // { value: "Lisa", done: false }- A:
yield getMembers(teams[i].members) - B:
yield* getMembers(teams[i].members) - C:
return getMembers(teams[i].members) - D:
return yield getMembers(teams[i].members)
Answer
In order to iterate over the members in each element in the teams array, we need to pass teams[i].members to the getMembers generator function. The generator function returns a generator object. In order to iterate over each element in this generator object, we need to use yield*.
If we would've written yield, return yield, or return, the entire generator function would've gotten returned the first time we called the next method.
const person = {
name: 'Lydia Hallie',
hobbies: ['coding'],
};
function addHobby(hobby, hobbies = person.hobbies) {
hobbies.push(hobby);
return hobbies;
}
addHobby('running', []);
addHobby('dancing');
addHobby('baking', person.hobbies);
console.log(person.hobbies);- A:
["coding"] - B:
["coding", "dancing"] - C:
["coding", "dancing", "baking"] - D:
["coding", "running", "dancing", "baking"]
Answer
The addHobby function receives two arguments, hobby and hobbies with the default value of the hobbies array on the person object.
First, we invoke the addHobby function, and pass "running" as the value for hobby and an empty array as the value for hobbies. Since we pass an empty array as the value for hobbies, "running" gets added to this empty array.
Then, we invoke the addHobby function, and pass "dancing" as the value for hobby. We didn't pass a value for hobbies, so it gets the default value, the hobbies property on the person object. We push the hobby dancing to the person.hobbies array.
Last, we invoke the addHobby function, and pass "baking" as the value for hobby, and the person.hobbies array as the value for hobbies. We push the hobby baking to the person.hobbies array.
After pushing dancing and baking, the value of person.hobbies is ["coding", "dancing", "baking"]
class Bird {
constructor() {
console.log("I'm a bird. ๐ฆข");
}
}
class Flamingo extends Bird {
constructor() {
console.log("I'm pink. ๐ธ");
super();
}
}
const pet = new Flamingo();- A:
I'm pink. ๐ธ - B:
I'm pink. ๐ธI'm a bird. ๐ฆข - C:
I'm a bird. ๐ฆขI'm pink. ๐ธ - D: Nothing, we didn't call any method
Answer
We create the variable pet which is an instance of the Flamingo class. When we instantiate this instance, the constructor on Flamingo gets called. First, "I'm pink. ๐ธ" gets logged, after which we call super(). super() calls the constructor of the parent class, Bird. The constructor in Bird gets called, and logs "I'm a bird. ๐ฆข".
const emojis = ['๐', '๐
๐ผ', '๐', 'โญ'];
/* 1 */ emojis.push('๐ฆ');
/* 2 */ emojis.splice(0, 2);
/* 3 */ emojis = [...emojis, '๐ฅ'];
/* 4 */ emojis.length = 0;- A: 1
- B: 1 and 2
- C: 3 and 4
- D: 3
Answer
The const keyword simply means we cannot redeclare the value of that variable, it's read-only. However, the value itself isn't immutable. The properties on the emojis array can be modified, for example by pushing new values, splicing them, or setting the length of the array to 0.
144. What do we need to add to the person object to get ["Lydia Hallie", 21] as the output of [...person]?
const person = {
name: "Lydia Hallie",
age: 21
}
[...person] // ["Lydia Hallie", 21]- A: Nothing, object are iterable by default
- B:
*[Symbol.iterator]() { for (let x in this) yield* this[x] } - C:
*[Symbol.iterator]() { yield* Object.values(this) } - D:
*[Symbol.iterator]() { for (let x in this) yield this }
Answer
Objects aren't iterable by default. An iterable is an iterable if the iterator protocol is present. We can add this manually by adding the iterator symbol [Symbol.iterator], which has to return a generator object, for example by making it a generator function *[Symbol.iterator]() {}. This generator function has to yield the Object.values of the person object if we want it to return the array ["Lydia Hallie", 21]: yield* Object.values(this).
let count = 0;
const nums = [0, 1, 2, 3];
nums.forEach(num => {
if (num) count += 1
})
console.log(count)- A: 1
- B: 2
- C: 3
- D: 4
Answer
The if condition within the forEach loop checks whether the value of num is truthy or falsy. Since the first number in the nums array is 0, a falsy value, the if statement's code block won't be executed. count only gets incremented for the other 3 numbers in the nums array, 1, 2 and 3. Since count gets incremented by 1 3 times, the value of count is 3.
function getFruit(fruits) {
console.log(fruits?.[1]?.[1])
}
getFruit([['๐', '๐'], ['๐']])
getFruit()
getFruit([['๐'], ['๐', '๐']])- A:
null,undefined, ๐ - B:
[],null, ๐ - C:
[],[], ๐ - D:
undefined,undefined, ๐
Answer
The ? allows us to optionally access deeper nested properties within objects. We're trying to log the item on index 1 within the subarray that's on index 1 of the fruits array. If the subarray on index 1 in the fruits array doesn't exist, it'll simply return undefined. If the subarray on index 1 in the fruits array exists, but this subarray doesn't have an item on its 1 index, it'll also return undefined.
First, we're trying to log the second item in the ['๐'] subarray of [['๐', '๐'], ['๐']]. This subarray only contains one item, which means there is no item on index 1, and returns undefined.
Then, we're invoking the getFruits function without passing a value as an argument, which means that fruits has a value of undefined by default. Since we're conditionally chaining the item on index 1 offruits, it returns undefined since this item on index 1 does not exist.
Lastly, we're trying to log the second item in the ['๐', '๐'] subarray of ['๐'], ['๐', '๐']. The item on index 1 within this subarray is ๐, which gets logged.
class Calc {
constructor() {
this.count = 0
}
increase() {
this.count ++
}
}
const calc = new Calc()
new Calc().increase()
console.log(calc.count)- A:
0 - B:
1 - C:
undefined - D:
ReferenceError
Answer
We set the variable calc equal to a new instance of the Calc class. Then, we instantiate a new instance of Calc, and invoke the increase method on this instance. Since the count property is within the constructor of the Calc class, the count property is not shared on the prototype of Calc. This means that the value of count has not been updated for the instance calc points to, count is still 0.
const user = {
email: "e@mail.com",
password: "12345"
}
const updateUser = ({ email, password }) => {
if (email) {
Object.assign(user, { email })
}
if (password) {
user.password = password
}
return user
}
const updatedUser = updateUser({ email: "new@email.com" })
console.log(updatedUser === user)- A:
false - B:
true - C:
TypeError - D:
ReferenceError
Answer
The updateUser function updates the values of the email and password properties on user, if their values are passed to the function, after which the function returns the user object. The returned value of the updateUser function is the user object, which means that the value of updatedUser is a reference to the same user object that user points to. updatedUser === user equals true.
const fruit = ['๐', '๐', '๐']
fruit.slice(0, 1)
fruit.splice(0, 1)
fruit.unshift('๐')
console.log(fruit)- A:
['๐', '๐', '๐'] - B:
['๐', '๐'] - C:
['๐', '๐', '๐'] - D:
['๐', '๐', '๐', '๐']
Answer
First, we invoke the slice method on the fruit array. The slice method does not modify the original array, but returns the value that it sliced off the array: the banana emoji.
Then, we invoke the splice method on the fruit array. The splice method does modify the original array, which means that the fruit array now consists of ['๐', '๐'].
At last, we invoke the unshift method on the fruit array, which modifies the original array by adding the provided value, โ๐โ in this case, as the first element in the array. The fruit array now consists of ['๐', '๐', '๐'].
const animals = {};
let dog = { emoji: '๐ถ' }
let cat = { emoji: '๐' }
animals[dog] = { ...dog, name: "Mara" }
animals[cat] = { ...cat, name: "Sara" }
console.log(animals[dog])- A:
{ emoji: "๐ถ", name: "Mara" } - B:
{ emoji: "๐", name: "Sara" } - C:
undefined - D:
ReferenceError
Answer
Object keys are converted to strings.
Since the value of dog is an object, animals[dog] actually means that weโre creating a new property called "object Object" equal to the new object. animals["object Object"] is now equal to { emoji: "๐ถ", name: "Mara"}.
cat is also an object, which means that animals[cat] actually means that weโre overwriting the value of animals["object Object"] with the new cat properties.
Logging animals[dog], or actually animals["object Object"] since converting the dog object to a string results "object Object", returns the { emoji: "๐", name: "Sara" }.
const user = {
email: "my@email.com",
updateEmail: email => {
this.email = email
}
}
user.updateEmail("new@email.com")
console.log(user.email)- A:
my@email.com - B:
new@email.com - C:
undefined - D:
ReferenceError
Answer
The updateEmail function is an arrow function, and is not bound to the user object. This means that the this keyword is not referring to the user object, but refers to the global scope in this case. The value of email within the user object does not get updated. When logging the value of user.email, the original value of my@email.com gets returned.
const promise1 = Promise.resolve('First')
const promise2 = Promise.resolve('Second')
const promise3 = Promise.reject('Third')
const promise4 = Promise.resolve('Fourth')
const runPromises = async () => {
const res1 = await Promise.all([promise1, promise2])
const res2 = await Promise.all([promise3, promise4])
return [res1, res2]
}
runPromises()
.then(res => console.log(res))
.catch(err => console.log(err))- A:
[['First', 'Second'], ['Fourth']] - B:
[['First', 'Second'], ['Third', 'Fourth']] - C:
[['First', 'Second']] - D:
'Third'
Answer
The Promise.all method runs the passed promises in parallel. If one promise fails, the Promise.all method rejects with the value of the rejected promise. In this case, promise3 rejected with the value "Third". Weโre catching the rejected value in the chained catch method on the runPromises invocation to catch any errors within the runPromises function. Only "Third" gets logged, since promise3 rejected with this value.
const keys = ["name", "age"]
const values = ["Lydia", 22]
const method = /* ?? */
Object[method](keys.map((_, i) => {
return [keys[i], values[i]]
})) // { name: "Lydia", age: 22 }- A:
entries - B:
values - C:
fromEntries - D:
forEach
Answer
The fromEntries method turns a 2d array into an object. The first element in each subarray will be the key, and the second element in each subarray will be the value. In this case, weโre mapping over the keys array, which returns an array which first element is the item on the key array on the current index, and the second element is the item of the values array on the current index.
This creates an array of subarrays containing the correct keys and values, which results in { name: "Lydia", age: 22 }
const createMember = ({ email, address = {}}) => {
const validEmail = /.+\@.+\..+/.test(email)
if (!validEmail) throw new Error("Valid email pls")
return {
email,
address: address ? address : null
}
}
const member = createMember({ email: "my@email.com" })
console.log(member)- A:
{ email: "my@email.com", address: null } - B:
{ email: "my@email.com" } - C:
{ email: "my@email.com", address: {} } - D:
{ email: "my@email.com", address: undefined }
Answer
The default value of address is an empty object {}. When we set the variable member equal to the object returned by the createMember function, we didn't pass a value for address, which means that the value of address is the default empty object {}. An empty object is a truthy value, which means that the condition of the address ? address : null conditional returns true. The value of address is the empty object {}.
let randomValue = { name: "Lydia" }
randomValue = 23
if (!typeof randomValue === "string") {
console.log("It's not a string!")
} else {
console.log("Yay it's a string!")
}- A:
It's not a string! - B:
Yay it's a string! - C:
TypeError - D:
undefined
Answer
The condition within the if statement checks whether the value of !typeof randomValue is equal to "string". The ! operator converts the value to a boolean value. If the value is truthy, the returned value will be false, if the value is falsy, the returned value will be true. In this case, the returned value of typeof randomValue is the truthy value "number", meaning that the value of !typeof randomValue is the boolean value false.
!typeof randomValue === "string" always returns false, since we're actually checking false === "string". Since the condition returned false, the code block of the else statement gets run, and Yay it's a string! gets logged.