Question

我正在尝试构建一个简单的monad计算器，但由于某种原因它不会编译。

{-# LANGUAGE GADTs #-}

module Main where

data Result a = Value a | Undefined deriving (Show, Eq)

data Calculator x a where
  Bind :: Calculator x a -> (a -> Calculator x b) -> Calculator x b
  Return :: a -> Calculator x a
  Add :: a -> a -> Calculator x a
  Div :: a -> a -> Calculator x a

instance Monad (Calculator x) where
  return = Return
  (>>=) = Bind

calculate :: Calculator (Result Integer) Integer
calculate = do
  value <- addr 1 2
  divr 1 value

addr :: a -> a -> Calculator (Result a) a
addr a b = Add a b

divr :: a -> a -> Calculator (Result a) a
divr a b = Div a b

eval :: (Integral a, Num a, Eq a) => Calculator (Result a) a -> Result a
eval (Add a b) = Value (a + b)
eval (Div a b)
  | b == 0    = Undefined
  | otherwise = Value $ a `div` b

eval (Bind m f) =
  case eval m of
    Value v -> eval $ f v
    _       -> undefined -- for now

我得到的错误就是这个

Main.hs:35:13:
    Could not deduce (a1 ~ a)
    from the context (Integral a, Num a, Eq a)
      bound by the type signature for
                 eval :: (Integral a, Num a, Eq a) =>
                         Calculator (Result a) a -> Result a
      at Main.hs:28:9-72
      ‘a1’ is a rigid type variable bound by
           a pattern with constructor
             Bind :: forall x b a.
                     Calculator x a -> (a -> Calculator x b) -> Calculator x b,
           in an equation for ‘eval’
           at Main.hs:34:7
      ‘a’ is a rigid type variable bound by
          the type signature for
            eval :: (Integral a, Num a, Eq a) =>
                    Calculator (Result a) a -> Result a
          at Main.hs:28:9
    Expected type: Calculator (Result a) a
      Actual type: Calculator (Result a) a1
    Relevant bindings include
      f :: a1 -> Calculator (Result a) a (bound at Main.hs:34:14)
      m :: Calculator (Result a) a1 (bound at Main.hs:34:12)
      eval :: Calculator (Result a) a -> Result a (bound at Main.hs:29:1)
    In the first argument of ‘eval’, namely ‘m’
    In the expression: eval m

错误发生在第三行到最后一行。谁知道为什么？

我正在使用ghci 7.8.4。我错过了什么？

Answer 1

首先，您的约束需要被推送到Calculator构造函数中。为了了解原因，让我们稍微重命名Bind签名中的类型变量：

data Calculator x a where
  Bind :: Calculator x b -> (b -> Calculator x a) -> Calculator x a

eval :: (Integral a, Num a, Eq a) => ...
eval (Bind m f) = ...

此处m的类型为Calculator x b，而a的约束（即(Integral a, Num a, Eq a)）不会对b施加任何限制，因此您没有希望在eval上递归调用m。

这个很容易解决：

data Calculator x a where
  Bind :: Calculator x a -> (a -> Calculator x b) -> Calculator x b
  Return :: a -> Calculator x a
  Add :: (Num a) => a -> a -> Calculator x a
  Div :: (Num a, Eq a, Integral a) => a -> a -> Calculator x a

然后eval不再需要a上的约束，因为它将通过相关构造函数上的模式匹配来提供。

另一个问题源于x的{{1}}类型参数。在

Calculator

请注意Bind :: Calculator x b -> (b -> Calculator x a) -> Calculator x a的选择保持不变。因此，当您在x上进行模式匹配时，您会得到

Bind m f :: Calculator (Result a) a

但是，m :: Calculator (Result a) b f :: b -> Calculator (Result a) a的类型只能实例化为eval或Calculator (Result a) a -> Result a，因此我们无法在Calculator (Result b) b -> Result b上递归应用它。

解决方案是在仿函数而不是类型上参数化m，因为我们可以输入Calculator：

eval :: Calculator Result a -> Result a

Monad错误（无法推断）

1 个答案: