Haskell：类型推理和函数组合 [英] Haskell: type inference and function composition

问题描述

这个问题是由这个 $ in-haskell-using-prelude / 1335530＃1335530＃ b $ b

  removeall = filter。 （/ =）
  

使用类型的铅笔和纸张进行处理过滤器，（/ =）和（。）

  removeall ::（Eq a）=> a  - > [a]  - > [a] 
  

这正是您期望根据合约得出的结果。然而，在GHCi 6.6中，我得到

  gchi> ：t removeall 
 removeall :: Integer  - > [整数]  - >除非我明确地指定类型（在这种情况下，它可以很好地工作），否则[整型] 
  

为什么Haskell为这个函数推断出这种特定的类型？为什么Haskell推断出这样一种特定类型的函数？特定类型的函数？

GHCi正在使用键入默认值，从一组可能值推断出更具体的类型。您可以通过禁用单态限制来轻松避免这种情况，

  Prelude> ：set -XNoMonomorphismRestriction 
 Prelude>让removeall = filter。 （/ =）
前奏> ：t removeall 
 removeall ::（Eq a）=> a  - > [a]  - > [a] 
  

This question was inspired by this answer to another question, indicating that you can remove every occurrence of an element from a list using a function defined as:

removeall = filter . (/=)

Working it out with pencil and paper from the types of filter, (/=) and (.), the function has a type of

removeall :: (Eq a) => a -> [a] -> [a]

which is exactly what you'd expect based on its contract. However, with GHCi 6.6, I get

gchi> :t removeall
removeall :: Integer -> [Integer] -> [Integer]

unless I specify the type explicitly (in which case it works fine). Why is Haskell inferring such a specific type for the function?

解决方案

Why is Haskell inferring such a specific type for the function?

GHCi is using type defaulting, to infer a more specific type from a set of possibles. You can avoid this easily by disabling the monomorphism restriction,

Prelude> :set -XNoMonomorphismRestriction
Prelude> let removeall = filter . (/=)
Prelude> :t removeall 
removeall :: (Eq a) => a -> [a] -> [a]

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