哈斯克尔（$）是一个魔术师？ [英] Haskell's ($) is a magic operator?

问题描述

假设我有以下功能：

  infixr 0< | 
 $ b $ { - ＃INLINE（< |）＃ - } 
（< |）::（a  - > b） - > a  - > b 
 f <| x = f x 
 
 foo :: a  - > （全部b.b→b）→> a 
 foo xf = fx 
  

以下不会输入check：

  ghci> foo 3 <| id 
 
无法匹配期望的类型`forall b。 b  - > b'
，实际类型为'a0  - > a0'
在'（< |）'的第二个参数中，即`id'
在表达式中：f 3 <| id 
在'it'的等式中：it = f 3 <| id 
  

然而， foo 3 $ id < （< |）的定义是（据我所知）与（$）的定义相同。 我几乎从基础库源中删除了定义，并将每个（$）的实例都更改为（< |）。 Compiler magic？

解决方案

是的，围绕（$）来处理impandicative类型。这是因为每个人都期望



  runST $ do 
 foo 
 bar 
 baz 
  

进行类型检查，但它不能正常运行。有关更多详情，请参阅此处（搜索 runST ），这个电子邮件和电子邮件。它的缺点是在类型检查器中实际上有一个特殊的规则，专门用于（$），这使得它能够解决一些常见的意外类型检查。

Say I have the following functions:

infixr 0 <|

{-# INLINE (<|) #-}
(<|) :: (a -> b) -> a -> b
f <| x = f x

foo :: a -> (forall b. b -> b) -> a
foo x f = f x

The following does not type check:

ghci> foo 3 <| id

Couldn't match expected type `forall b. b -> b'
            with actual type `a0 -> a0'
In the second argument of `(<|)', namely `id'
In the expression: f 3 <| id
In an equation for `it': it = f 3 <| id

However, foo 3 $ id does.

The definition of (<|) is (as far as I know) identical to the definition of ($). I pretty much ripped out the definition from the base library sources, and changed every instance of ($) to (<|). Compiler magic?

解决方案

Yes, there's a small amount of compiler magic around ($) to handle impredicative types. It was introduced because everyone expects

runST $ do
  foo
  bar
  baz

to typecheck, but it cannot normally. For more details see here (search runST), this email, and this email. The short of it is that there's actually a special rule in the type-checker specifically for ($) which gives it the ability to resolve the common case of impredicative types.

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