Haskell：确定函数的函数的函数？ [英] Haskell: Function to determine the arity of functions?

问题描述

是否可以编写一个函数 arity :: a - > Integer 来确定任意函数的形式，例如

 > arity map 
 2 
> arity foldr 
 3 
> arity id 
 1 
> arityhello
 0 
  

？

OverlappingInstances ：

   { - ＃LANGUAGE FlexibleInstances，OverlappingInstances＃ - } 
 
类Arity f其中
 arity :: f  - > Int 
 
实例Arity x其中
 arity _ = 0 
 
实例Arity f => Arity（（ - >）af）其中
 arity f = 1 + arity（f undefined）
  

更新发现问题。您需要为多态函数指定非多态类型：

  arity（foldr ::（a  - > Int  - > Int）> Int> [a]  - > Int）
  

不知道如何解决这个问题呢。

Upd2 由于Sjoerd Visscher在下面评论说：您必须指定非多态类型，因为答案取决于您选择哪种类型。

Is it possible to write a function arity :: a -> Integer to determine the arity of arbitrary functions, such that

> arity map
2
> arity foldr
3
> arity id
1
> arity "hello"
0

?

解决方案

It's easy with OverlappingInstances:

{-# LANGUAGE FlexibleInstances, OverlappingInstances #-}

class Arity f where
  arity :: f -> Int

instance Arity x where
  arity _ = 0

instance Arity f => Arity ((->) a f) where
  arity f = 1 + arity (f undefined) 

Upd Found problem. You need to specify non-polymorphic type for polymorphic functions:

arity (foldr :: (a -> Int -> Int) -> Int -> [a] -> Int)

Don't know how to solve this yet.

Upd2 as Sjoerd Visscher commented below "you have to specify a non-polymorphic type, as the answer depends on which type you choose".

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