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查看 ghci 的输出：

Prelude> :t Data.Map.lookup
Data.Map.lookup :: Ord k => k -> Data.Map.Map k a -> Maybe a
Prelude> :t flip Data.Map.lookup
flip Data.Map.lookup :: Ord a => Data.Map.Map a a1 -> a -> Maybe a1
Prelude> let look = flip Data.Map.lookup
Loading package array-0.3.0.2 ... linking ... done.
Loading package containers-0.4.0.0 ... linking ... done.
Prelude> :t look
look :: Data.Map.Map () a -> () -> Maybe a

为什么look推断类型与类型不同flip Data.Map.lookup？

给你一些背景。最初我有一个小程序，并试图弄清楚它为什么会产生编译器错误：

import qualified Data.Map as M

type A = String
type B = String
data C = C1 | C2 | C3
     deriving (Eq, Ord)
type D = String

z :: A -> M.Map A B -> M.Map B C -> M.Map C D -> Maybe D
z a aToB bToC cToD = look aToB a >>= look bToC >>= look cToD
  where look = flip M.lookup

Ghci 的反应：

Prelude> :load main.hs
[1 of 1] Compiling Main             ( main.hs, interpreted )
Failed, modules loaded: none.

main.hs:10:52:
    Couldn't match expected type `C' with actual type `[Char]'
    Expected type: C -> Maybe D
      Actual type: A -> Maybe a0
    In the return type of a call of `look'
    In the second argument of `(>>=)', namely `look cToD'

我发现这个变体编译得很好（类型定义是一样的）：

x :: A -> M.Map A B -> M.Map B C -> Maybe C
x a aToB bToC = look aToB a >>= look bToC
  where look = flip M.lookup

y :: A -> M.Map A B -> M.Map B C -> M.Map C D -> Maybe D
y a aToB bToC cToD = (x a aToB bToC) >>= look cToD
  where look = flip M.lookup

经过一些实验后发现，如果我look明确输入 type of - 第一个版本也编译得很好：

z :: A -> M.Map A B -> M.Map B C -> M.Map C D -> Maybe D
z a aToB bToC cToD = look aToB a >>= look bToC >>= look cToD
  where look :: (Ord a) => M.Map a b -> a -> Maybe b
        look = flip M.lookup

这引出了我的第一个问题。

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1 回答 1

7

默认情况下，顶级绑定是非多态的，除非给出明确的类型说明符；这被称为“单态限制”。k由于没有给出类型说明符，GHC 必须在定义函数时选择一种实例化方式。它碰巧选择了k = ()。

这背后的想法是，多态性会通过在最终编译的代码中引入大量 vtable 调用来损害性能；除非另有明确说明，否则通过强制在编译时解决这些问题，可以避免这种开销。这一决定颇具争议。GHC 支持扩展以完全禁用单态限制，通过传递-XNoMonomorphismRestriction.

于 2011-07-31T02:56:44.673 回答