Question

在GHC中编译此程序时：

import Control.Monad

f x = let
  g y = let
    h z = liftM not x
    in h 0
  in g 0

我收到错误：

test.hs:5:21:
    Could not deduce (m ~ m1)
    from the context (Monad m)
      bound by the inferred type of f :: Monad m => m Bool -> m Bool
      at test.hs:(3,1)-(7,8)
    or from (m Bool ~ m1 Bool, Monad m1)
      bound by the inferred type of
               h :: (m Bool ~ m1 Bool, Monad m1) => t1 -> m1 Bool
      at test.hs:5:5-21
      `m' is a rigid type variable bound by
          the inferred type of f :: Monad m => m Bool -> m Bool
          at test.hs:3:1
      `m1' is a rigid type variable bound by
           the inferred type of
           h :: (m Bool ~ m1 Bool, Monad m1) => t1 -> m1 Bool
           at test.hs:5:5
    Expected type: m1 Bool
      Actual type: m Bool
    In the second argument of `liftM', namely `x'
    In the expression: liftM not x
    In an equation for `h': h z = liftM not x

为什么呢？此外，为f（f :: Monad m => m Bool -> m Bool）提供显式类型签名会使错误消失。但这与Haskell根据错误消息自动推断f的类型完全相同！

Answer 1

实际上，这非常简单。推断的let - 绑定变量的类型被隐式推广到类型方案，所以你的方式有一个量词。广义类型h是：

h :: forall a m. (Monad m) => a -> m Bool

f的广义类型是：

f :: forall m. (Monad m) => m Bool -> m Bool

他们不是m。如果你这样写，你会得到基本相同的错误：

f :: (Monad m) => m Bool -> m Bool
f x = let
  g y = let
    h :: (Monad m) => a -> m Bool
    h z = liftM not x
    in h 0
  in g 0

你可以通过启用“范围类型变量”扩展来修复它：

{-# LANGUAGE ScopedTypeVariables #-}

f :: forall m. (Monad m) => m Bool -> m Bool
f x = let
  g y = let
    h :: a -> m Bool
    h z = liftM not x
    in h 0
  in g 0

或通过禁用let - 使用“单态本地绑定”扩展名MonoLocalBinds进行泛化。

无法推断（m~m1）

1 个答案: