Solved–Assignment 4: Nano Overview –Solution

Description

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][/src/Language/Nano/Lexer.x]

2. [Parser.y][/src/Language/Nano/Parser.y]

3. [Eval.hs][/src/Language/Nano/Lexer.x]

and

+ [tests/Test.hs](/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:

“`haskell

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](/tests/Test.hs) contains a very small suite

of tests which gives you a flavor of of these tests.

When you run

“`shell

$ 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, push your code to gitlab and turn in the commit ID on canvas.

## 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`.

“`haskell

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`.

“`haskell

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:

“`haskell

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

“`haskell

let x = 3 in x + x

“`

is represented by

“`haskell

ELet “x” (EInt 3)

(EBin Plus (EVar “x”) (EVar “x”))

“`

2. Anonymous Functions definitions

“`haskell

\x -> x + 1

“`

is represented by

“`haskell

ELam “x” (EBin Plus (EVar “x”) (EInt 1))

“`

3. Function applications (“calls”)

“`haskell

f x

“`

is represented by

“`haskell

EApp (EVar “f”) (EVar “x”)

“`

4. (Recursive) Named Functions

“`haskell

let f = \ x -> f x in

f 5

“`

is represented by

“`haskell

ELet “f” (ELam “x” (EApp (EVar “f”) (EVar “x”)))

(EApp (Var “f”) (EInt 5))

“`

Values

We will represent Nano **values**, i.e. the results

of evaluation, using the following datatype

“`haskell

data Value

= VInt Int

| VBool Bool

| VClos Env Id Expr

| VNil

| VPair 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:

“`haskell

type Env = [(Id, Value)]

“`

Intuitively, the Nano integer value `4` and boolean value

`True` are represented respectively as `VInt 4` and `VBool True`.

– `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:

“`haskell

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

“`haskell

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:

“`haskell

throw (Error (“unbound variable: ” ++ x))

“`

When you are done you should get the following behavior:

“`haskell

>>> 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

“`haskell

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:

“`haskell

>>> eval env0 (EBin Minus (EBin Plus “x” “y”) (EBin Plus “z” “z1”))

0

>>> eval env0 “p”

*** Exception: Error {errMsg = “unbound variable: p”}

“`

(b) 20 points

Next, add support for the binary operators

“`haskell

data Binop = …

| Eq | Ne | Lt | Le | And | Or

“`

This will require using the new value type `Bool`

“`haskell

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.

“`haskell

throw (Error “type error”)

“`

When you are done, you should see the following behavior

“`haskell

>>> 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:

“`haskell

>>> let e1 = EIf (EBin Lt “z1” “x”) (EBin Ne “y” “z”) (EBool False)

>>> eval env0 e1

True

>>> let e2 = EIf (EBin Eq “z1” “x”) (EBin Le “y” “z”) (EBin Le “z” “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.

“`haskell

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:

“`haskell

>>> let e1 = EBin Plus “x” “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.

“`haskell

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.

“`haskell

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:

“`haskell

>>> eval [] (EApp (ELam “x” (EBin Plus “x” “x”)) (EInt 3))

6

>>> let e3 = ELet “h” (ELam “y” (EBin Plus “x” “y”)) (EApp “f” “h”)

>>> let e2 = ELet “x” (EInt 100) e3

>>> let e1 = ELet “f” (ELam “g” (ELet “x” (EInt 0) (EApp “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:

“`haskell

— >>> :{

— eval [] (ELet “fac” (ELam “n” (EIf (EBin Eq “n” (EInt 0))

— (EInt 1)

— (EBin Mul “n” (EApp “fac” (EBin Minus “n” (EInt 1))))))

— (EApp “fac” (EInt 10)))

— :}

— 3628800

“`

(f) 40 points

Finally, extend your program to support operations on lists.

“`haskell

data Binop = …

| Cons

data Expr = …

| ENil

data Value = …

| VNil

| VPair 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

“`haskell

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

>>> execExpr el

(1 : (2 : []))

>>> execExpr (EApp “head” el)

1

>>> execExpr (EApp “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 `let` 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.

Once you have implemented this functionality, you should get the

following behavior:

“`haskell

>>> 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:

“`haskell

>>> 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` |

| `in` | `IN` |

| `\` | `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.

Once you have implemented this functionality

you should get the following behavior prompt:

“`haskell

>>> 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` |

| `&&` | `AND` |

| `||` | `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:

“`haskell

>>> 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 the following tokens to the lexer and parser.

|String | Token |

|:——|:———-|

| `(` | `LPAREN` |

| `)` | `RPAREN` |

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:

“`haskell

>>> 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”))))))

“`

(d) 35 points

Restructure your parser to give binary operators the

following precedence and associativity. This will

likely require that you add additional rules to your

parser, or see how to

[add precedence and associativity to Happy](https://www.haskell.org/happy/doc/html/sec-Precedences.html)

**Operators Precedence Order**

+ (Highest) Fun Application

+ `*`

+ `+`, `-`

+ `==`, `/=`, `<`, `<=`

+ `&&`

+ (Lowest) `||`

**Precedence** Function application having higher precedence than

multiplications, and multiplication higher than addition

means that `”1+f x*3″` should be parsed as if it were

`”1+((f x)*3)”`.

**Associativity** All Operators are *Left associative*

means that `”1-2-3-4″` should be parsed as if it were

`”((1-2)-3)-4″`, and `”f x y z”` should be parsed as

if it were `”((f x) y) z”`.

Once you have implemented this functionality and recompiled,

you should get the following behavior:

“`haskell

>>> 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”)))

“`

(e) 15 points

Add the following tokens to the lexer and parser.

String Token

——– ——–

`[` `LBRAC`

`]` `RBRAC`

`,` `COMMA`

`:` `COLON`

Add rules to your lexer and parser to support parsing lists.

`”[a,b,c,d,e,f,g]”` should be parsed as if it were

`”a:b:c:d:e:f:g:[]”`. 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.

Once you have implemented this functionality you should get

the following behavior

“`haskell

>>> 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))

“`