14

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

我收到一个错误：

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

4

1 回答 1

5

实际上，这很简单。-bound 变量的推断类型let隐式推广到类型方案，因此您的方式中有一个量词。的广义类型h是：

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

的广义类型f是：

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

他们不一样m。如果你写这个，你会得到基本上相同的错误：

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

您可以通过启用“作用域类型变量”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

或者通过let使用“单态局部绑定”扩展禁用 -generalisation MonoLocalBinds，.

于 2013-07-20T20:07:09.677 回答