制作类型:添加数据类型和不匹配类型的评估器 [英] making types: Add data type and evaluator that is of an unmatched type

问题描述

我添加了|点亮Int和|将Exp与数据评估一起添加到如下所示的数据类型中.但是我收到错误消息无法将预期的类型"Var"与实际类型"Int"匹配".

I have added | Lit Int and | Add Exp Exp to the data type as seen below, along with the evaluation. However I get an error "Couldn't match expected type ‘Var’ with actual type ‘Int’".

data Exp = V Var
    | B Bool
    | L Exp
    | A Exp Exp
    | Lit Int
    | Add Exp Exp
data Var = VZ |VS Var

eval:: Exp -> Var
eval (Lit n)     = n
eval (Add e1 e2) = eval e1 + eval e2

如何将Int和Add与评估一起添加到数据类型中，但按原样维护以下代码.这可能吗?

How can I add Int and Add to the data type, along with the evaluation, but maintain the following code as is. Is this possible?

data Exp = V Var
    | B Bool
    | L Exp
    | A Exp Exp
data Var = VZ |VS Var

---

我已经添加了一个实例来解决此问题，如下所示，但是现在我出现了错误实例声明中不允许出现模式绑定(简单变量除外):Add e1 e2 =(Lit e1)+(Lit e2)

---

I have added an instance to resolve this, as seen below, but now I have the error "Pattern bindings (except simple variables) not allowed in instance declaration: Add e1 e2 = (Lit e1) + (Lit e2)":

{-# LANGUAGE TypeSynonymInstances, FlexibleInstances #-}

data Exp = V Var
    | B Bool
    | L Exp
    | A Exp Exp
    | Lit Int
    | Add Exp Exp
data Var = VZ |VS Var

eval:: Exp -> Var
eval (Lit n)     = n
eval (Add e1 e2) = eval e1 + eval e2

instance Num Var where
    Lit e = e

instance Num Var where
    Add e1 e2 = (Lit e1) + (Lit e2)

推荐答案

您似乎在此过程中感到困惑.我相信 Var 表示 De Bruijn中的免费变量符号.您缺少值的编码.这可能看起来像是模糊的

You seem to have gotten mixed up along the way. I believe Var represents a free variable in De Bruijn notation. You're missing an encoding of values. This could look something vaguely like

data Value
  = IntVal !Int
  | BoolVal !Bool
  | Closure { environment :: [Value]
            , expr :: Exp }

现在，您需要 eval :: Exp->值以评估表达式.还有很多工作.

Now you'll want eval :: Exp -> Value to evaluate expressions. There's lots of work left.

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