将函数应用于任意长的参数列表

发布于 2024-08-31 12:31:59 字数 345 浏览 13 评论 0原文

我想创建一个函数 apply，它接受带有任意数量参数的函数以及整数列表，并返回函数的结果（其中列表中的每个整数都是按顺序排列的参数。

我在想类似的事情：

apply :: ([Int] -> Int) -> [Int] -> Int
apply f x:xs = apply (f x) xs
apply f [] = f

但我知道这行不通，因为类型签名是错误的 - 该函数不接受整数列表，它只接受一些 int 参数

此外，当我到达基本情况时，要应用 f 参数。实际上应该是一个整数，无论如何都违反了类型签名。

有谁知道如何处理此类问题？

原文

I want to create a function apply that takes a function with an arbitrary amount of arguments as well as a list of integers, and returns the result of the function (Where each integer in the list is an argument in order.

I was thinking something like:

apply :: ([Int] -> Int) -> [Int] -> Int
apply f x:xs = apply (f x) xs
apply f [] = f

But I know this won't work because the type signature is wrong - the function doesn't take a list of ints, it just takes some amount of int arguments.

Additionally, when I get to the base case the f argument to apply should actually be an integer, violating the type signature anyway.

Does anyone know how to deal with this sort of problem?

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神回复 2024-09-07 12:31:59

我想创建一个函数 apply，它接受一个带有任意数量参数的函数以及一个整数列表，

为什么要这样做？也许你的参数结构应该作为数据结构传递，但到目前为止你已经过度限制了问题以确保它不会产生惯用的 Haskell 解决方案。

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耳根太软 2024-09-07 12:31:59

你可以用一些奇特的类型类来做到这一点

{-# LANGUAGE FlexibleInstances #-}
-- for ApplyType (Int -> r)

class ApplyType t where
    apply :: t -> [Int] -> Int

instance ApplyType Int where
    apply f _ = f

instance (ApplyType r) => ApplyType (Int -> r) where
    apply f (x:xs) = apply (f x) xs

main :: IO ()
main = do print $ apply ((+) :: Int->Int->Int) [1, 2]
          print $ apply ((\x y z w -> x*y - z`div`w) :: Int->Int->Int->Int->Int) [3,5,8,2]

You can do it with some fancy type classes

{-# LANGUAGE FlexibleInstances #-}
-- for ApplyType (Int -> r)

class ApplyType t where
    apply :: t -> [Int] -> Int

instance ApplyType Int where
    apply f _ = f

instance (ApplyType r) => ApplyType (Int -> r) where
    apply f (x:xs) = apply (f x) xs

main :: IO ()
main = do print $ apply ((+) :: Int->Int->Int) [1, 2]
          print $ apply ((\x y z w -> x*y - z`div`w) :: Int->Int->Int->Int->Int) [3,5,8,2]

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