关于这两个haskell符号的问题以及它们的作用：$和\ [英] Question about the two haskell symbols and what they do: $ and \

问题描述

有人可以用简单的语言向我解释那两个操作员做的事：

  $ 
 
 \ 
  

解决方案

\ 不是一个运算符，它是字面语法的一部分。更确切地说，它是两个文字语法的一部分：它表示一个lambda文字，它在字符串文字中用作转义字符。

运算符 $ 在前奏中被定义为

 （$）:: （a  - > b） - > a  - > b 
f $ x = fx 
  

换句话说， em>与空格相同，即只是简单的函数应用。 ，而函数应用程序是左关联的并且具有高优先级（实际上最高）， $ 是右关联的，并且具有低优先级。

这样可以在你将< f 这样的链应用于时省略括号。应用于 x 的应用于 h ，其中 > $ 运算符会写成像



  f（g（hx））
  

但用运算符可以写成


  f $ g $ hx 
  

如果您想通过函数应用程序运算符本身作为另一个函数的参数。比如说，你有函数列表和值列表，你想将列表中的每个函数应用到另一个列表中的对应值：



  zipWith（$）fs xs 
  




Can someone explain to me with simple words what those two operators do:

$

\

解决方案

\ is not an operator, it is part of the literal syntax. More precisely, it is part of two literal syntaxes: it denotes a lambda literal and it serves as an escape character in string literals.

The operator $ is defined in the prelude as

($) :: (a -> b) -> a -> b
f $ x = f x

In other words, it does exactly the same thing as whitespace does, namely just plain function application. However, while function application is left-associative and has high precedence (the highest, in fact), $ is right-associative and has low precedence.

This allows you to omit parentheses when you have chains like "f applied to g applied to h applied to x", which without the $ operator would be written like

f (g (h x))

but with the operator can be written as

f $ g $ h x

It is also useful if you want to pass the function application operator itself as an argument to another function. Say, you have list of functions and a list of values and you want to apply every function in the list to the corresponding value in the other list:

zipWith ($) fs xs

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