Question

我不明白为什么demonbind1的定义会产生一些编译器错误。它看起来像一个愚蠢的翻转但不知何故..

{-# LANGUAGE GADTs #-}
{-# LANGUAGE RankNTypes, ScopedTypeVariables, TypeOperators, TypeFamilies,ImpredicativeTypes   #-}

type a :-> b = forall i . a i -> b i

class IFunctor f where imap :: (a :-> b) -> (f a :-> f b)

class (IFunctor m) => IMonad m where
    skip :: a :-> m a
    bind :: (a :-> m b) -> (m a :-> m b)


-- Conor McBride's "demonic bind"
(?>=) :: forall m a b i. (IFunctor m, IMonad m) => m a i -> (a :-> m b) -> m b i
(?>=) =  
  let
    -- OK 
    demonbind0 = flip (bind :: forall i. (forall j. a j -> m b j)  -> m a i -> m b i ) 
    -- KO - see error below
    demonbind1 = flip  bind :: forall i. m a i -> (forall j. a j -> m b j)  ->  m b i

    -- So i have to write this
    demonbind2 :: forall i. (m a i ->  (a :-> m b)  ->  m b i )
    demonbind2 mai ti = (bind ti) mai 
  in demonbind2

错误是

Couldn't match type ‘a j0 -> m b j0’ …
              with ‘forall i2. a i2 -> m b i2’
    Expected type: (a j0 -> m b j0) -> m a i1 -> m b i1
      Actual type: a :-> m b -> m a i1 -> m b i1
 In the first argument of ‘flip’, namely ‘bind’
    In the expression:
        flip bind :: forall i. m a i -> (forall j. a j -> m b j) -> m b i

Answer 1

有点令人惊讶的是，ImpredicativeTypes在GHC 8.0的开发快照中似乎没有那么糟糕！这编译没有错误：

(?>=) :: (IFunctor m, IMonad m) => m a i -> (a :-> m b) -> m b i
(?>=) = flip bind

我想知道是什么改变解决了这个问题。

RankN多态和kleisli箭头的离谱财富

1 个答案: