haskellmutual-recursionevaluatortying-the-knotmonadfix

Mutually recursive evaluator in Haskell


Update: I've added an answer that describes my final solution (hint: the single Expr data type wasn't sufficient).


I'm writing an evaluator for a little expression language, but I'm stuck on the LetRec construct.

This is the language:

type Var = String
type Binds = [(Var, Expr)]

data Expr
  = Var     Var
  | Lam     Var    Expr
  | App     Expr   Expr
  | Con     Int
  | Sub     Expr   Expr
  | If      Expr   Expr  Expr
  | Let     Var    Expr  Expr
  | LetRec  Binds  Expr
  deriving (Show, Eq)

And this this the evaluator so far:

data Value
  = ValInt   Int
  | ValFun   Env   Var  Expr
  deriving (Show, Eq)

type Env = [(Var, Value)]

eval :: Env -> Expr -> Either String Value
eval env (Var x)       = maybe (throwError  $ x ++ " not found")
                               return
                               (lookup x env)
eval env (Lam x e)     = return $ ValFun env x e
eval env (App e1 e2)   = do
                         v1 <- eval env e1
                         v2 <- eval env e2
                         case v1 of
                           ValFun env1 x e -> eval ((x, v2):env1) e
                           _ -> throwError "First arg to App not a function"
eval _   (Con x)       = return $ ValInt x
eval env (Sub e1 e2)   = do
                         v1 <- eval env e1
                         v2 <- eval env e2
                         case (v1, v2) of
                           (ValInt x, ValInt y) -> return $ ValInt (x - y)
                           _ -> throwError "Both args to Sub must be ints"
eval env (If p t f)    = do 
                         v1 <- eval env p
                         case v1 of
                           ValInt x -> if x /= 0
                                       then eval env t
                                       else eval env f
                           _ -> throwError "First arg of If must be an int"
eval env (Let x e1 e2) = do
                         v1 <- eval env e1
                         eval ((x, v1):env) e2
eval env (LetRec bs e) = do
                         env' <- evalBinds
                         eval env' e
  where
    evalBinds = mfix $ \env' -> do
      env'' <- mapM (\(x, e') -> eval env' e' >>= \v -> return (x, v)) bs
      return $ nub (env'' ++ env)

This is my test function I want to evaluate:

test3 :: Expr
test3 = LetRec [ ("even", Lam "x" (If (Var "x")
                                      (Var "odd" `App` (Var "x" `Sub` Con 1))
                                      (Con 1)
                                  ))
               , ("odd",  Lam "x" (If (Var "x")
                                      (Var "even" `App` (Var "x" `Sub` Con 1))
                                      (Con 0)
                                  ))
               ]
               (Var "even" `App` Con 5)

EDIT:

Based on Travis' answer and Luke's comment, I've updated my code to use the MonadFix instance for the Error monad. The previous example works fine now! However, the example bellow doesn't work correctly:

test4 :: Expr
test4 = LetRec [ ("x", Con 3)
               , ("y", Var "x")
               ]
               (Con 0)

When evaluating this, the evaluator loops, and nothing happens. I'm guessing I've made something a bit too strict here, but I'm not sure what it is. Am I violating one of the MonadFix laws?


Solution

  • When Haskell throws a fit, that's usually an indication that you have not thought clearly about a core issue of your problem. In this case, the question is: which evaluation model do you want to use for your language? Call-by-value or call-by-need?

    Your representation of environments as [(Var,Value)] suggests that you want to use call-by-value, since every Expr is evaluated to a Value right away before storing it in the environment. But letrec does not go well with that, and your second example shows!

    Furthermore, note that the evaluation model of the host language (Haskell) will interfere with the evaluation model of the language you want to implement; in fact, that's what you are currently making use of for your examples: despite their purpose, your Values are not evaluated to weak head normal form.

    Unless you have a clear picture of the evaluation model of your little expression language, you won't make much progress on letrec or on the error checking facilities.

    Edit: For an example specification of letrec in a call-by-value language, have a look at the Ocaml Manual. On the simplest level, they only allow right-hand sides that are lambda expressions, i.e. things that are syntactically known to be values.