kennethpham / 04-nano

Starter code for hw4

Assignment 4: Nano (270 points)

Overview

The overall objective of this assignment is to fully understand the notions of

• lexing,
• parsing,
• scoping,
• binding,
• environments and closures,

by implementing an interpreter for a subset of Haskell.

No individual function requires more than 15-25 lines, so if you're answer is longer, you can be sure that you need to rethink your solution.

The assignment is in the files:

1. Lexer.x
2. Parser.y
3. Eval.hs

and

• tests/Test.hs has some sample tests, and testing code that you will use to check your assignments before submitting.

You should only need to modify the parts of the files which say:

error "TBD: ..."

with suitable Haskell implementations.

---

Note: Start early! Lexing and Parsing are new tools, which may take a while to grok.

---

Assignment Testing and Evaluation

Most of the points, will be awarded automatically, by evaluating your functions against a given test suite.

Tests.hs contains a very small suite of tests which gives you a flavor of of these tests. When you run

$ stack test

Your last lines should have

All N tests passed (...)
OVERALL SCORE = ... / ...

or

K out of N tests failed
OVERALL SCORE = ... / ...

If your output does not have one of the above your code will receive a zero

If for some problem, you cannot get the code to compile, leave it as is with the error ... with your partial solution enclosed below as a comment.

The other lines will give you a readout for each test. You are encouraged to try to understand the testing code, but you will not be graded on this.

Submission Instructions

To submit your code, just do:

$ make turnin

turnin will provide you with a confirmation of the submission process; make sure that the size of the file indicated by turnin matches the size of your file. See the ACS Web page on turnin for more information on the operation of the program.

or alternatively, just git push you code to your github classroom repository.

Data Structures and Overview

In this assignment, you will build an interpreter for a subset of Haskell called Nano. The following data types (in Types.hs) are used to represent the different elements of the language.

Binary Operators

Nano uses the following binary operators encoded within the interpreter as values of type Binop.

data Binop
  = Plus
  | Minus
  | Mul
  | Div
  | Eq
  | Ne
  | Lt
  | Le
  | And
  | Or
  | Cons

Expressions

All Nano programs correspond to expressions each of which will be represented within your interpreter by Haskell values of type Expr.

data Expr
  = EInt  Int
  | EBool Bool
  | ENil
  | EVar Id
  | EBin Binop Expr Expr
  | EIf  Expr Expr  Expr
  | ELet Id   Expr  Expr
  | EApp Expr Expr
  | ELam Id   Expr
  deriving (Eq)

where Id is just a type alias for String used to represent variable names:

type Id = String

The following lists some Nano expressions, and the value of type Expr used to represent the expression inside your interpreter.

1. Let-bindings

let x = 3 in x + x

is represented by

ELet "x" (EInt 3)
  (EBin Plus (EVar "x") (EVar "x"))

2. Anonymous Functions definitions

\x -> x + 1

is represented by

ELam "x" (EBin Plus (EVar "x") (EInt 1))

3. Function applications ("calls")

f x

is represented by

EApp (EVar "f") (EVar "x")

4. (Recursive) Named Functions

let f = \ x -> f x in
  f 5

is represented by

ELet "f" (ELam "x" (EApp (EVar "f") (EVar "x")))
  (EApp (EVar "f") (EInt 5))

Values

We will represent Nano values, i.e. the results of evaluation, using the following datatype

data Value
  = VInt  Int
  | VBool Bool
  | VClos Env Id Expr
  | VNil
  | VCons Value Value
  | VPrim (Value -> Value)

where an Env is simply a dictionary: a list of pairs of variable names and the values they are bound to:

type Env = [(Id, Value)]

Intuitively, the Nano integer value 4 and boolean value True are represented respectively as VInt 4 and VBool True. The more interesting case is for closures that correspond to function values (see lecture notes).

• VClos env "x" e represents a function with argument "x" and body-expression e that was defined in an environment env.

Problem 1: Nano Interpreter (Eval.hs)

In this problem, you will implement an interpreter for Nano.

(a) 25 points

First consider the (restricted subsets of) types described below:

data Binop = Plus | Minus | Mul

data Expr  = EInt Int
           | EVar Id
           | EBin Binop Expr Expr

data Value = VInt Int

That is,

• An expression is either an Int constant, a variable, or a binary operator applied to two sub-expressions.

• A value is an integer, and an environment is a list of pairs of variable names and values.

Write a Haskell function

lookupId :: Id -> Env -> Value

where lookupId x env returns the most recent binding for the variable x (i.e. the first from the left) in the list representing the environment. If no such value is found, you should throw an error:

throw (Error ("unbound variable: " ++ x))

When you are done you should get the following behavior:

>>> lookupId "z1" env0
0

>>> lookupId "x" env0
1

>>> lookupId "y" env0
2

>>> lookupId "mickey" env0
*** Exception: Error {errMsg = "unbound variable: mickey"}

Next, use lookupId to write a function

eval :: Env -> Expr -> Value

such that eval env e evaluates the Nano expression e in the environment env (i.e. uses env for the values of the free variables in e), and throws an Error "unbound variable" if the expression contains a free variable that is not bound in env.

Once you have implemented this functionality and recompiled, you should get the following behavior:

>>> eval env0 (EBin Minus (EBin Plus (EVar "x") (EVar "y")) (EBin Plus (EVar "z") (EVar "z1")))
0

>>> eval env0 (EVar "p")
*** Exception: Error {errMsg = "unbound variable: p"}

(b) 20 points

Next, add support for the binary operators

data Binop = ...
           | Eq | Ne | Lt | Le | And | Or

This will require using the new value type Bool

data Value = ...
           | VBool Bool

• The operators Eq and Ne should work if both operands are VInt values, or if both operands are VBool values.

• The operators Lt and Le are only defined for VInt values, and && and || are only defined for VBool values.

• Other pairs of arguments are invalid and you should throw a suitable error.

throw (Error "type error")

When you are done, you should see the following behavior

>>> eval []  (EBin Le (EInt 2) (EInt 3))
True

>>> eval []  (EBin Eq (EInt 2) (EInt 3))
False

>>> eval []  (EBin Lt (EInt 2) (EBool True))
*** Exception: Error {errMsg = "type error: binop"}

Also note that, so long as you error message is appropriate, you will receive points. We will not be checking for an exact error message. However, it should contain the substring 'type error:'.

Next, implement the evaluation of EIf p t f expressions.

1. First, evaluate the p; if p does not evaluate to a VBool value, then your evaluator should throw (Error "type error"),

2. If p evaluates to the true value then the expression t should be evaluated and returned as the value of the entire If expression,

3. Instead, if p evaluates to the false value, then f should be evaluated and that result should be returned.

Once you have implemented this functionality, you should get the following behavior:

>>> let e1 = EIf (EBin Lt (EVar "z1") (EVar "x")) (EBin Ne (EVar "y") (EVar "z")) (EBool False)
>>> eval env0 e1
True

>>> let e2 = EIf (EBin Eq (EVar "z1") (EVar "x")) (EBin Le (EVar "y") (EVar "z")) (EBin Le (EVar "z") (EVar "y"))
>>> eval env0 e2
False

(c) 25 points

Now consider the extended the types as shown below which includes the let-in expressions which introduce local bindings.

data Expr
  = ...
  | ELet Id   Expr  Expr

The expression ELet x e1 e2 should be evaluated as the Haskell expression let x = e1 in e2.

Once you have implemented this functionality and recompiled, you should get the following behavior:

>>> let e1 = EBin Plus (EVar "x")  (EVar "y")
>>> let e2 = ELet "x" (EInt 1) (ELet "y" (EInt 2) e1)
>>> eval [] e2
3

(d) 25 points

Next, extend the evaluator so it includes the expressions corresponding to function definitions and applications.

data Expr
  = ...
	| ELam Id   Expr
	| EApp Expr Expr

In the above,

• ELam x e corresponds to the function defined \x -> e, and

• EApp e1 e2 corresponds to the Haskell expression e1 e2 (i.e. applying the argument e2 to the function e1).

To evaluate functions, you will need to extend the set of values yielded by your evaluator to include closures.

data Value
  = ...
	| VClos Env Id Expr

For now, assume the functions are not recursive.

However, functions do have values represented by the VClos env x e where

• env is the environment at the point where that function was declared,
• x is the formal parameter, and
• e the body expression of the function.

Extend your implementation of eval by adding the appropriate cases for the new type constructors. Once you have implemented this functionality and recompiled, you should get the following behavior:

>>> eval [] (EApp (ELam "x" (EBin Plus (EVar "x") (EVar "x"))) (EInt 3))
6

>>> let e3 = ELet "h" (ELam "y" (EBin Plus (EVar "x") (EVar "y"))) (EApp (EVar "f") (EVar "h"))
>>> let e2 = ELet "x" (EInt 100) e3
>>> let e1 = ELet "f" (ELam "g" (ELet "x" (EInt 0) (EApp (EVar "g") (EInt 2)))) e2
>>> eval [] e1
102

(e) 30 points

Make the above work for recursively defined functions. Once you have implemented this functionality, you should get the following behavior:

-- >>> :{
-- eval [] (ELet "fac" (ELam "n" (EIf (EBin Eq (EVar "n") (EInt 0))
--                                  (EInt 1)
--                                  (EBin Mul (EVar "n") (EApp (EVar "fac") (EBin Minus (EVar "n") (EInt 1))))))
--             (EApp (EVar "fac") (EInt 10)))
-- :}
-- 3628800

(f) 40 points

Finally, extend your program to support operations on lists.

data Binop = ...
           | Cons

data Expr = ...
          | ENil

data Value = ...
           | VNil
           | VCons Value Value

In addition to the changes to the data types, add support for two functions head and tail which do what the corresponding Haskell functions do. Once you have implemented this functionality and recompiled, you should get the following behavior

>>> let el = EBin Cons (EInt 1) (EBin Cons (EInt 2) ENil)

>>> execExpr el
(1 : (2 : []))

>>> execExpr (EApp (EVar "head") el)
1

>>> execExpr (EApp (EVar "tail") el)
(2 : [])

The constructor VPrim will come in handy here.

Problem 2: Nano Lexer (Lexer.x) and Parser (Parser.y)

The goal of this problem is to write a lexer and parser for Nano using the tools Alex and Happy. (Google those terms for more information about them.) In each subproblem, we will increase the complexity of the expressions parsed by your implementation.

(a) 15 points

We will begin by making our parser recognize some of the simplest Nano expressions: constants and variables.

Begin with Lexer.x using the given rules for keywords in and && as an inspiration, fill in the rules for

• TRUE and FALSE which should correspond to the string literals True and False.

• ID which has a single String argument, which holds the name of the variable (identifier) represented by the token. An identifier is a letter (capital or lowercase) followed by zero or more letters or digits.

• NUM which has a single Int argument, which holds the value of the numeric literal, which corresponds to a sequence of one or more digits.

---

Note Be careful when implementing the regular expressions for in Lexer.x, as an incorrect expression can cause your program to leak memory pretty quickly. You'll want to either watch your memory usage and be ready to kill the test, or use a memory limiting tool like this one: timeout.

---

Once you have implemented this functionality, you should get the following behavior:

>>> parseTokens "True"
Right [TRUE (AlexPn 0 1 1)]

>>> parseTokens "True False 12345 foo bar baz"
Right [TRUE (AlexPn 0 1 1),FALSE (AlexPn 5 1 6),NUM (AlexPn 11 1 12) 12345,ID (AlexPn 17 1 18) "foo",ID (AlexPn 21 1 22) "bar",ID (AlexPn 25 1 26) "baz"]

The AlexPn n l c denote the position in the string where the token was parsed. For example, the FALSE is at character 5, line 1 and column 6.

Now proceed to Parser.y.

Add rules to the parser so that True, False, integers, and identifiers are parsed into suitable Expr values.

Once you have implemented this functionality, you should get the following behavior:

>>> parse "True"
EBool True

>>> parse "False"
EBool False

>>> parse "123"
EInt 123

>>> parse "foo"
EVar "foo"

(b) 15 points

Add the following tokens to the lexer and parser.

String	Token
let	LET
=	EQB
\	LAM
->	ARROW
if	IF
then	THEN
else	ELSE

These should be parsed to ELet, ELam, and EIf expressions, that is,

• a let expression should have the form let <id> = <expr> in <expr>, or the form let <id> <ids> = <expr> in <expr>,
• a function expression should have the form \ <id> -> <expr> and
• an if expression should be if <expr> then <expr> else <expr>.

Here <id> denotes any identifier from part (a), <ids> denotes a sequence of one or many space-separated <id>s, and <expr> denotes any expression from part (a), or any let / fun / if expression.

---

Note

The order of rules in your lexer definitions matters: earlier patterns will match first, then later patters, much like the function cases in Haskell.

---

Once you have implemented this functionality you should get the following behavior prompt:

>>> parseTokens "let foo = \\x -> if y then z else w in foo"

Right [LET (AlexPn 0 1 1),ID (AlexPn 4 1 5) "foo",EQB (AlexPn 8 1 9),
       LAM (AlexPn 10 1 11),ID (AlexPn 11 1 12) "x",ARROW (AlexPn 13 1 14),
       IF (AlexPn 16 1 17),ID (AlexPn 19 1 20) "y",THEN (AlexPn 21 1 22),
       ID (AlexPn 26 1 27) "z",ELSE (AlexPn 28 1 29),ID (AlexPn 33 1 34) "w",
       IN (AlexPn 35 1 36),ID (AlexPn 38 1 39) "foo"]

>>> parse "let foo = \\x -> if y then z else w in foo"
ELet "foo" (ELam "x" (EIf (EVar "y") (EVar "z") (EVar "w"))) (EVar "foo")

>>> parse "let foo x = if y then z else w in foo"
ELet "foo" (ELam "x" (EIf (EVar "y") (EVar "z") (EVar "w"))) (EVar "foo")

(c) 15 points

Add the following tokens to the lexer and parser.

String	Token
+	PLUS
-	MINUS
*	MUL
<	LESS
<=	LEQ
==	EQL
/=	NEQ
||	OR

Add all of these as binary operators to your parser. Each should result in a EBin expression with the corresponding binop. The arguments to these binary operators may be any expressions. (You don't need to worry about types: 3 + True || 7 is allowed as far as the parser is concerned.)

Once you have implemented this functionality and recompiled, you should get the following behavior:

>>> parseTokens "+ - * || < <= = && /="
Right [PLUS (AlexPn 0 1 1),MINUS (AlexPn 2 1 3),
       MUL (AlexPn 4 1 5),OR (AlexPn 6 1 7),
       LESS (AlexPn 9 1 10),LEQ (AlexPn 11 1 12),
       EQB (AlexPn 14 1 15),AND (AlexPn 16 1 17),
       NEQ (AlexPn 19 1 20)]

>>> parse "x + y"
EBin Plus (EVar "x") (EVar "y")

>>> parse "if x <= 4 then a || b else a && b"
EIf (EBin Le (EVar "x") (EInt 4)) (EBin Or (EVar "a") (EVar "b")) (EBin And (EVar "a") (EVar "b"))

>>> parse "if 4 <= z then 1 - z else 4 * z"
EIf (EBin Le (EInt 4) (EVar "z")) (EBin Minus (EInt 1) (EVar "z")) (EBin Mul (EInt 4) (EVar "z"))

>>> parse "let a = 6 * 2 in a /= 11"
ELet "a" (EBin Mul (EInt 6) (EInt 2)) (EBin Ne (EVar "a") (EInt 11))

(d) 10 points

Add rules to your parser to allow parenthesized expressions. In addition, add a rule to your parser for function application. Recall that function application is simply "<expr> <expr>" which corresponds to calling the (function corresponding to the) left expression with the (argument corresponding to the) right expression.

Once you have implemented this functionality and recompiled, you should get the following behavior:

>>> parseTokens "() (  )"
Right [LPAREN (AlexPn 0 1 1),RPAREN (AlexPn 1 1 2),LPAREN (AlexPn 3 1 4),RPAREN (AlexPn 6 1 7)]

>>> parse "f x"
EApp (EVar "f") (EVar "x")

>>> parse "(\\ x -> x + x) (3 * 3)"
EApp (ELam "x" (EBin Plus (EVar "x") (EVar "x"))) (EBin Mul (EInt 3) (EInt 3))

>>> parse "(((add3 (x)) y) z)"
EApp (EApp (EApp (EVar "add3") (EVar "x")) (EVar "y")) (EVar "z")

>>> parse <$> readFile "tests/input/t1.hs"
EBin Mul (EBin Plus (EInt 2) (EInt 3)) (EBin Plus (EInt 4) (EInt 5))

>>> parse <$> readFile "tests/input/t2.hs"
ELet "z" (EInt 3) (ELet "y" (EInt 2) (ELet "x" (EInt 1) (ELet "z1" (EInt 0) (EBin Minus (EBin Plus (EVar "x") (EVar "y")) (EBin Plus (EVar "z") (EVar "z1"))))))

(e) 35 points

Restructure your parser to give binary operators the following precedence and associativity. This will likely require that you both add additional rules to your parser and use Happy's precedence syntax

Operators Precedence Order

• (Highest) Fun Application
• *
• +, -
• ==, /=, <, <=
• &&
• ||
• ->
• (Lowest) =, if, then, else, in

Precedence Function application having higher precedence than multiplications, and multiplication higher than addition, so "1+f x*3" should be parsed as if it were "1+((f x)*3)".

Associativity All arithmetic and logical operators, as well as function application are left associative, so "1-2-3-4" should be parsed as "((1-2)-3)-4", and "f x y z" should be parsed as "((f x) y) z".

Once you have implemented this functionality and recompiled, you should get the following behavior:

>>> parse "1-2-3"
EBin Minus (EBin Minus (EInt 1) (EInt 2)) (EInt 3)
>>> parse "1+a&&b||c+d*e-f-g x"
EBin Or (EBin And (EBin Plus (EInt 1) (EVar "a")) (EVar "b")) (EBin Minus (EBin Minus (EBin Plus (EVar "c") (EBin Mul (EVar "d") (EVar "e"))) (EVar "f")) (EApp (EVar "g") (EVar "x")))

(f) 15 points

Add the following tokens to the lexer

String	Token
[	LBRAC
]	RBRAC

Add rules to your parser to support parsing lists. "[a,b,c]" should be parsed as if it were "((a):(b):(c):[])". The : operator should have higher priority than the comparison functions (==, <= etc.), and lower priority than + and -. In addition, : should be right associative. "[]" should be parsed as ENil, and : should be treated as any other binary operator.

Also, implement equality for lists in Eval.hs.

Once you have implemented this functionality you should get the following behavior

>>> parse "1:3:5:[]"
EBin Cons (EInt 1) (EBin Cons (EInt 3) (EBin Cons (EInt 5) ENil))

>>> parse "[1,3,5]"
EBin Cons (EInt 1) (EBin Cons (EInt 3) (EBin Cons (EInt 5) ENil))