Haskell编写你的（$）函数的版本 [英] Haskell write your version of a ($) function

问题描述

有人可以帮我编写我自己版本的（$）函数吗？
这是我的尝试，不起作用。

  f函数（x：xs）= f（x：xs）
  

解决方案

首先，您为什么认为您需要：或 foldr 在这里？这些对列表是排他性的。 $ 与列表无关。

f function（x）= f x）更喜欢它，但是您似乎对函数定义 f 作为参数，反之亦然。事实上， f函数（x）= f（x）表示：

  f = \ function x  - > f x 
  

即您将 f 定义为一个函数，该函数接受一个名为的函数（这是从未实际使用的）和另一个参数 x 然后它将你正在定义的函数应用到这里......这种递归结合在Haskell中实际上是可能的（有时非常有用），但在这种情况下，它没有任何意义。

实际上你想写的内容要简单得多：

  f $ x = fx 
  

请注意<$ c因为它由非字母符号组成，因此是一个 infix ，因此分析方式不同：

<$> pre> （$）= \ fx - > fx


也就是说， $ 和一个参数并将该函数应用于参数;就是这样。

你也可以通过使用反引号将字母名称分解为中缀来实现。

  f`function` x = fx 
  

...或者简单地

pre $ code>函数fx = fx

Could someone help me with writing my own version of the ($) function? This is my attempt which does not work.

f function (x:xs) = f (x:xs)

解决方案

First, why do you think you need : or foldr here? These are exclusive to lists. $ has nothing to do with lists.

f function (x) = f (x) is more like it, however it seems you're confused about whether you defining function with f as an argument, or vice versa. In fact, f function (x) = f (x) means this:

f = \function x -> f x

i.e. you're defining f as a function which takes an argument called function (that is never actually used) and another argument x to which it then applies the very function you're defining right here... this kind of recursive knot-tying is actually possible in Haskell (and sometimes quite useful), but in this case it doesn't make any sense.

What you actually want to write is much simpler:

f $ x = f x

note that $, because it consists of a non-letter symbol, is an infix and therefore parsed differently:

($) = \f x -> f x

Which means, $ takes a function and an argument and applies the function to the argument; that's it.

You could also achieve this with a name with letters by using backticks to have it parsed as an infix:

f `function` x = f x

...or simply

function f x = f x

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