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Home/ Questions/Q 9245893
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Editorial Team
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Editorial Team
Asked: June 18, 20262026-06-18T09:22:01+00:00 2026-06-18T09:22:01+00:00

I am new to Haskell. I am getting this Not in scope: data constructor

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I am new to Haskell.
I am getting this “Not in scope: data constructor `String'” in almost every line of the code that contains “String”

Please take a look at my code and let me know what I am doing wrong.
I really appreciate your help, thanks.

import Data.Maybe

data Op = Add | Sub | Mul | Div | And | Or | Not | Eq | Less | Great
    deriving (Eq, Show)

data Exp = Literal Value
     | Primitive Op [Exp]
     | Variable String
     | If Exp Exp Exp
     | Let [(String, Exp)] Exp
    deriving (Show, Eq)

data Value = Number Int
       | Bool Bool
    deriving (Eq, Show)

type Env = [(String, Value)]

eval :: Env -> Exp -> Value
eval e (Literal v) = v
eval e (Variable x) = fromJust (lookup x e)
--22
prim :: op -> [Value] -> Value
prim Add [Number a, Number b] = Number (a+b)
prim And [Bool a, Bool b] = Bool (a && b)
prim Sub [Number a, Number b] = Number (a-b)
prim Mul [Number a, Number b] = Number (a*b)
prim Div [Number a, Number b] = Number (a/b)
prim Or [Bool a, Bool b] = Bool (a || b)
prim Not [Bool a] = Bool (not a)
prim Eq [Number a, Number b] = Bool (a == b)
prim Eq [String a, String b] = Bool (a == b) 
prim Less [Number a, Number b] = Bool (a < b)
prim Less [String a, String b] = Bool (a < b)
prim Great [Number a, Number b] = Bool (a > b)
prim Great [String a, String b] = Bool (a > b) 
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  1. Editorial Team
    Editorial Team
    2026-06-18T09:22:02+00:00Added an answer on June 18, 2026 at 9:22 am

    Looks like you forgot to add String to your Value datatype?

    data Value = Number Int
           | Bool Bool
           | String String
        deriving (Eq, Show)
    
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