Question

这是玩具示例.hs：

{-# LANGUAGE ImpredicativeTypes #-}

import Control.Arrow

data From = From (forall a. Arrow a => a Int Char -> a [Int] String)

data Fine = Fine (forall a. Arrow a => a Int Char -> a () String)

data Broken = Broken (Maybe (forall a. Arrow a => a Int Char -> a () String))

fine :: From -> Fine
fine (From f) = Fine g
  where g :: forall a. Arrow a => a Int Char -> a () String
        g x = f x <<< arr (const [1..5])

broken :: From -> Broken
broken (From f) = Broken (Just g) -- line 17
  where g :: forall a. Arrow a => a Int Char -> a () String
        g x = f x <<< arr (const [1..5])

这就是ghci的想法：

GHCi, version 7.0.3: http://www.haskell.org/ghc/  :? for help
Loading package ghc-prim ... linking ... done.
Loading package integer-gmp ... linking ... done.
Loading package base ... linking ... done.
Loading package ffi-1.0 ... linking ... done.
Prelude> :l toy-example.hs 
[1 of 1] Compiling Main             ( toy-example.hs, interpreted )

toy-example.hs:17:32:
    Couldn't match expected type `forall (a :: * -> * -> *).
                                  Arrow a =>
                                  a Int Char -> a () String'
                with actual type `a0 Int Char -> a0 () String'
    In the first argument of `Just', namely `g'
    In the first argument of `Broken', namely `(Just g)'
    In the expression: Broken (Just g)
Failed, modules loaded: none.

为什么在fine没有broken时进行类型检查？

如何让broken进行类型检查？

（在我的实际代码中，如果必须的话，我可以将类型参数a添加到Broken，而不是在构造函数中对其进行普遍量化，但我想尽可能避免这种情况。。）

修改：如果我将Broken的定义更改为

data Broken = Broken (forall a. Arrow a => Maybe (a Int Char -> a () String))

然后broken类型检查。耶！

但如果我再添加以下功能

munge :: Broken -> String
munge (Broken Nothing) = "something"  -- line 23
munge (Broken (Just f)) = f chr ()

然后我收到错误消息

toy-example.hs:23:15:
    Ambiguous type variable `a0' in the constraint:
      (Arrow a0) arising from a pattern
    Probable fix: add a type signature that fixes these type variable(s)
    In the pattern: Nothing
    In the pattern: Broken Nothing
    In an equation for `munge': munge (Broken Nothing) = "something"

如何让munge同时进行类型检查？

第二次编辑：在我的真实程序中，我已将Broken (Maybe ...)构造函数替换为BrokenNothing和BrokenJust ...构造函数（已有其他构造函数），但我我很好奇模式匹配在这种情况下应该如何工作。

Answer 1

ImpredicativeTypes让你处于非常不稳定的境地，无论如何都会从GHC版本变为版本 - 他们正在努力寻找一种适当平衡功能，易用性和易于实施的不可信度的表述。

在这种特殊情况下，尝试将量化类型放在Maybe内，这是一种未明确定义的数据类型，这种方式非常棘手，因此我建议使用自定义构造函数相反，如你所说。

我认为你可以通过在RHS上重新解构munge的参数来修复Broken，当时使用它的类型将是已知，例如：

munge (Broken x@(Just _)) = fromJust x chr ()

但是，这很丑陋。

为什么不这个类型检查呢？

1 个答案: