quitschibo / java8-lambda-expression-cheat-sheet

java8-lambda-expression-cheat-sheet

General

a -> a + 1; // take a as parameter and return a + 1

(b, c) -> b + c; // take (b, c) as parameter and return b + c

// take (c, d, e, f) as parameter and return the sum as return value.
// The lambda implements LambdaInterface and has the name myLambda
LambdaInterface myLambda = (c, d, e, f) -> c + d + e + f; 

Collections

Iterate a list: list.forEach(i -> System.out.print("Element: " + i);

Iterate a list with filtering:

list.stream()
	.filter(i -> i % 3 == 0) 
	.forEach(i -> System.out.print("Element: " + i));

Filtering and mapping:

list.stream()
	.filter(i -> i % 3 == 0)
	.map(i -> new Double(i))

Filtering, mapping and collecting:

ArrayList<Double> doubleList = list.stream()
	.filter(i -> i % 3 == 0)
	.map(i -> new Double(i))
	.collect(Collectors.toCollection(ArrayList::new));

Parallel vs Sequential

You can run the mapping and filtering sequential:

// if the ints are ordered in list, the result will be ordered, too
list.stream()
	.sequential()
	.filter(i -> i % 3 == 0)
	.map(i -> new Double(i))
	.forEach(i -> System.out.println("Element: " + i));

You can run it parallel:

// even if the ints are ordered in list, the result will NOT be ordered
list.stream()
	.parallel()
	.filter(i -> i % 3 == 0)
	.map(i -> new Double(i))
	.forEach(i -> System.out.println("Element: " + i));

BUT:

// if the list is ordered, the result will be ordered too, even if it is processed parallel.
// rule of thumb: if you expect it to be ordered, it will be ordered (lists, arrays, etc.)
ArrayList<Double> result = list.stream()
	.parallel()
	.filter(i -> i % 3 == 0)
	.map(i -> new Double(i))
	.collect(Collectors.toCollection(ArrayList::new));

Example

  /**
   * The interface of my lambda
   */
  interface LambdaInterface {
	  String doSomething(String message);
  }
  
  /**
   * This is the method which should use my lambda expression
   */
  public static void lamdaUsingMethod(String messageToLog, LambdaInterface lambda) {
	  System.out.print(lambda.doSomething(messageToLog));
  }
  
  /**
   * The main method which run the testprogram
   */
  public static void main(String[] args) {
	  // here we build the lambda expression
	  LambdaInterface errorMessage = (message) -> "Error: " + message;
    
    // here we use the lambda to so something
	  lamdaUsingMethod("This is my testmessage", errorMessage);
  }