Haskell：带有匿名/ lambda函数的函数组合

问题描述

  t1 =（reverse）在学习功能性编程考试时，我遇到了以前的测试中的以下问题： 。2。words。\ _  - > name）！ 
  

任务是编写语句的输出。变量名称是以Smith，John的形式编写的学生姓名。如果我将声明输入到WinHugs中，我会得到以下输出：

  [John，Smith，] 
  

我明白什么是函数反转，运营商连接它们。我不明白这里发生了什么：

  \ _  - >名称
  

什么是斜杠，下划线和箭头？另外，引号中的惊叹号是做什么的？ （没有？）

解决方案

这是一个lambda函数，它放弃其（仅）参数（即！）并产生 name 。

作为另一个lambda例子，下面是一个lambda函数，它的参数是正方形：

  \x  - > x * x 
  

\ 是

_ 表示关于我们不关心其名称的变量。

$ b

- > 将lambda函数的参数与用于指定结果的表达式分开。

While studying for a Functional Programming exam, I came across the following question from a previous test:

t1 = (reverse . take 2 . words . \ _ -> name)"!"

The task is to write the output of the statement. The variable name refers to the student's name, written in the form "Smith, John". If I enter the statement into WinHugs, I get the following output:

["John","Smith,"]

I understand what the functions reverse, take and words are doing and I understand how the . operator connects them. What I don't understand is what is happening here:

\ _ -> name

What are the slash, underscore and "arrow" for? Also, what does the exclamation point in quotation marks do? (nothing?)

解决方案

It's a lambda function that discards its (only) argument (i.e. "!") and yields name.

As another lambda example, the following would be a lambda function that squares its argument:

\x -> x * x

The \ is the notation used to introduce a lambda function.

The _ means "variable about whose name we do not care".

The -> separates the lambda function's arguments from the expression used to specify its result.

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