Haskell given loop issue

Question

I came across the same problem, but maybe I can give a little more context in what is happening for my situation, and maybe this can help.

Creating an interpreter using typedef:

interp :: Env -> Expr -> M Val

which processes loops in this format:

for ( var = expr to expr ) ( expr )

The data constructor is defined as follows:

data Val =
   ValInt Int
 | ValBool Bool
 | ValFun (Val -> M Val)
 | ValRecFun (Val -> Val -> M Val)
 | ValRef Loc
 | ValNil

And an extended environment defined as:

extendEnv :: Identifier -> Val -> Env -> Env
extendEnv var val (Env bs) = Env ((var,val):bs)

Here where I am:

interp env (For x e1 e2 e3)       = do
                                      (ValInt v1) <- interp env e1
                                      (ValInt v2) <- interp env e2
                                      if (v1 < v2)
                                        then 
                                            let nenv = extendEnv x e1 env in do
                                                interp nenv e3
                                                interp env (For x e1 e2 e3)
                                        else return ValNil

Obviously, I don’t want to pass “e1” to the recursive call of the for loop, but rather, the evaluated variable “v1” is incremented ... but I cannot figure out how to pass the correct expression “v1” to it. Is this enough to help a little? :)

* UPDATE *

, doLoop, . , , , "env" doLoop, .

interp env (For x e1 e2 e3)= do
                          (ValInt v1) <- interp env e1
                          (ValInt v2) <- interp env e2
                     return doLoop x v1 v2 env e3

doLoop :: Identifier -> Int -> Int -> Env-> Expr -> M Val
doLoop x v1 v2 env e3 = 
                 if v1 > v2 then return ValNil
                 else
                    let nenv = extendEnv x (ValInt v1) env in
                    interp nenv e3
                    doLoop x (ValInt (v1+1)) v2 nenv e3

UPDATE

, , :

return doLoop x v1 v2 env e3 

M Val' against my inferred type Expr → M Val '.

- ?

Answer 1

, "e1", , , "v1" ... , "v1". , ?:)

"" e1 v1, 1 v1 . makeConstExpr , Expr.

interp env (For var start end body) = do 
 (ValInt s) <- interp env start
 (ValInt e) <- interp env end 
 if s <= e 
   then
       let envn = extendEnv var (ValInt s) env in do
     interp envn body
     interp env (For var (makeConstExpr (s + 1)) end body) 
   else return ValNil

(, , , ...)

, , -. , , , , , .

: , , nevn env interp env (For var (ValInt (s + 1)) end body).

Answer 2

, . ( Ints, Vals, , parens env ):

doLoop :: Identifier -> Int -> Int -> Env -> Expr -> M Val
doLoop x v1 v2 env e3 = 
                 if v1 > v2 then return ValNil
                 else
                    let nenv = extendEnv x (ValInt v1) env in 
                    interp nenv e3
                    doLoop x (ValInt (v1+1)) v2 env e3