07-variables

Module 07: Variables

  • Write your team names here:

In this module, we will explore how to support a language with variables.

We’ll start with the familiar Arith language.

{-# LANGUAGE GADTSyntax #-}

import Prelude hiding ((<$>), (<$), (<*>), (<*), (*>))
import Parsing

data Arith where
  Lit :: Integer -> Arith
  Bin :: Op -> Arith -> Arith -> Arith
  deriving (Show)

data Op where
  Plus  :: Op
  Minus :: Op
  Times :: Op
  deriving (Show, Eq)

interpArith :: Arith -> Integer
interpArith (Lit i)           = i
interpArith (Bin Plus e1 e2)  = interpArith e1 + interpArith e2
interpArith (Bin Minus e1 e2) = interpArith e1 - interpArith e2
interpArith (Bin Times e1 e2) = interpArith e1 * interpArith e2

lexer :: TokenParser u
lexer = makeTokenParser $ emptyDef
  { reservedNames = ["let", "in"] }
    -- tell the lexer that "let" and "in" are reserved keywords
    -- which may not be used as variable names

parens :: Parser a -> Parser a
parens     = getParens lexer

reservedOp :: String -> Parser ()
reservedOp = getReservedOp lexer

reserved :: String -> Parser ()
reserved = getReserved lexer

integer :: Parser Integer
integer    = getInteger lexer

whiteSpace :: Parser ()
whiteSpace = getWhiteSpace lexer

identifier :: Parser String
identifier = getIdentifier lexer

parseArithAtom :: Parser Arith
parseArithAtom = (Lit <$> integer) <|> parens parseArith

parseArith :: Parser Arith
parseArith = buildExpressionParser table parseArithAtom
  where
    table = [ [ Infix (Bin Times <$ reservedOp "*") AssocLeft ]
            , [ Infix (Bin Plus  <$ reservedOp "+") AssocLeft
              , Infix (Bin Minus <$ reservedOp "-") AssocLeft
              ]
            ]

arith :: Parser Arith
arith = whiteSpace *> parseArith <* eof

eval :: String -> Maybe Integer
eval s = case parse arith s of
  Left _  -> Nothing
  Right e -> Just (interpArith e)

We will now add variables to this language. In particular, we’re going to add let-expressions, which look something like this:

 >>> let x = 4*12 in x*x + 3-x
 2259

This locally defines x as a name for the value of 4*12 within the expression x*x + 3-x. Substituting 48 for each occurrence of x and evaluating the result yields the final value of 2259.

Haskell has let-expressions too, so you can try typing the above expression at a GHCi prompt.

Syntax

  • Choose a driver and write their name here:

We need to make two changes to the syntax of the language.

  • Arith expressions may now contain variables, represented as Strings.

  • Arith expressions may now contain let-expressions, which have the concrete syntax 'let' <var> '=' <arith> 'in' <arith>

    where <var> is a variable name (a String) and the two occurrences of <arith> are Arith expressions.

  • Add two new constructors to the definition of the Arith data type to represent these two new syntactic forms. (Of course, since Arith represents abstract syntax, you do not need to worry about storing specific pieces of concrete syntax such as the words let or in; for a let expression you need only store the variable name and two Arith expressions.)

  • Modify the parser to parse the given concrete syntax. A few hints:

    • You can use the provided identifier :: Parser String to parse a variable name.
    • You can use the provided reserved :: String -> Parser () to parse the keywords let and in. Notice how these keywords are specified as “reserved names” in the lexer, which means they may not be used as variable names.
    • You should extend the definition of parseArithAtom with two new cases, corresponding to the two new constructors of Arith. To keep the code a bit cleaner, you may want to create a new auxiliary definition called parseLet to parse a let-expression; or you can simply inline it into the definition of parseArithAtom.

  • Be sure to test your parser on some sample inputs! For example, you might try things like

    • x+y-foo
    • let x = 4*48 in x*x + x-3
    • let x = 3 in let y = x*x in let z = y*y*x*x in z*z*y*y*x*x
    • (let baz = 19 in baz*baz) - (let foo = 12 in foo*foo)

    but you should make up your own examples to try as well. All of these should parse successfully.

Semantics

  • ROTATE ROLES and write the name of the new driver here:

Now we will extend the interpreter appropriately. The presence of variables introduces two new wrinkles to our interpreter:

  1. The interpreter will need to keep track of the current values of variables, so that when we encounter a variable we can substitute its value.
  2. Variables introduce the possibility of runtime errors. For example, let x = 2 in x + y will generate a runtime error, since y is undefined. (Later we will see how to make this a compile-time error instead.)

For now, you should deal with undefined variables simply by calling the error function with an appropriate error message, which will make the interpreter crash. In later modules we will explore better ways to handle runtime errors.

To keep track of the values of variables, we will use a mapping from variable names to values, called an environment. To represent such a mapping, we can use the Data.Map module, which works similarly to Python dictionaries or Java’s Map interface. Add the following line to the imports at the top of this file:

import qualified Data.Map as M

You can now refer to things from Data.Map by prefixing them with M., for example, M.empty, M.insert, M.lookup. (The reason for prefixing with M like this is that otherwise there would be conflicts with functions from the Prelude.)

  • Define an environment as a map from variable names to values: type Env = M.Map String Integer

  • Now add an extra Env parameter to interpArith which keeps track of the current values of variables.

  • Extend interpArith to interpret variables and let-expressions. You will probably find the M.insert and M.lookup functions useful. See the Data.Map documentation for information on these (and other) functions.

  • (Optional, just for fun): try to write down an expression which evaluates to an integer which is longer than the expression, that is, the integer has more digits than the number of characters in the original expression. What is the shortest expression you can write with this property? This was not possible with the previous version of Arith but it is possible now that we have let and variables.