由于（-1）被视为负数，等于（+1）的相减是多少？ [英] What is the equivalent to (+1) for the subtraction, since (-1) is seen as a negative number?

问题描述

可能存在重复：

柯鲁里减法

我开始了我的第一个不是来自教程的haskell项目，并且

我有以下代码：

  moveUp y = modifyMVar_ y $ return。 （+1）
 moveDn y = modifyMVar_ y $ return。 （-1）
  

我花了一些时间来理解为什么我的代码不能编译：使用（-1）被视为负数。括号中的减号不会帮助它，因为它将它作为前缀并使其成为第一个参数。

简而言之，这是什么免费版？

  dec :: Num a => a  - > a 
 dec x = x  -  1 
  

解决方案

我相信你需要方便地命名的减去函数，这正是您发现的原因：

  subtract :: Num a => a  - > a  - > a 
  

与相同flip（ - ） 。

因为在Haskell语法中特别处理 - ，所以（ - e）不是节，而是前缀否定的应用。但是，（减去exp）等同于不允许的部分。

如果你可以使用 flip（ - ）作为 Prelude 文档提及。但那是...有点丑。

Possible Duplicate:
Currying subtraction

I started my first haskell project that is not from a tutorial, and of course I stumble on the simplest things.

I have the following code:

moveUp y = modifyMVar_ y $ return . (+1)
moveDn y = modifyMVar_ y $ return . (-1)

It took me some time to understand why my code wouldn't compile: I had used (-1) which is seen as negative one. Bracketting the minus doesn't help as it prefixes it and makes 1 its first parameter.

In short, what is the point free version of this?

dec :: Num a => a -> a
dec x = x - 1

解决方案

I believe you want the conveniently-named subtract function, which exists for exactly the reason you've discovered:

subtract :: Num a => a -> a -> a

the same as flip (-).

Because - is treated specially in the Haskell grammar, (- e) is not a section, but an application of prefix negation. However, (subtract exp) is equivalent to the disallowed section.

If you wanted to write it pointfree without using a function like subtract, you could use flip (-), as the Prelude documentation mentions. But that's... kinda ugly.

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