不了解此haskell代码中的内容 [英] Dont understand whats on in this haskell code

问题描述

我有一些haskell代码，我正在试图按自己的方式工作，但我不了解其中发生了什么.

I have some haskell code Im trying to work my way thourgh but I dont have understand what is going in it.

type Bag a = a -> Int

emptyB :: Bag a
emptyB = \e -> 0

countB :: Eq a => Bag a -> a -> Int
countB b e = b e

我知道Bag类型是一个接受通用对象并返回Int的函数，而countB基本上是Bag的包装，该包装获取该Bag中通用对象的数量.但是我真的不明白那件事.我如何修改袋子里的东西?还是包本身?从我看来，添加到书包中将是

I understand that the Bag type is a function that takes in a generic object and returns a Int and countB is basically a wrapper for Bag that gets the number of generic objects in that Bag. But I dont really understand anything past that. How do I modify whats in the bag? Or the bag itself? From what I figure adding to the bag would be something like

addB :: Eq a => Bag a -> a -> Bag a
addB bag num = bag (num+bag) 

但是，当add函数要求返回bag时，这将返回一个int值.谁能向我解释这是如何工作的?

But this returns a int, when the add function requires a bag be returned. Can anyone explain to me how this works?

推荐答案

术语和讨论

type Bag a = a -> Int

此处 Bag 不是对象.它只是一种类型- a->的别名.Int .如果您有一个 a 类型的值，它将计算并返回一个 Int 类型的值.就是这样.没有袋子，没有可以添加东西的结构.最好不要把它叫做袋子.

Here Bag is not an object. It is just a type - an alias for a -> Int. If you have a value of type a it will compute and return a value of type Int. That's it. There is no Bag, no structure to which you can add things. It would be better to not even call this a Bag.

emptyB :: Bag a
emptyB = \e -> 0

从任何类型到常数0的函数.

A function from any type to the constant number zero.

countB :: Eq a => Bag a -> a -> Int
countB b e = b e

简而言之，这只是函数应用程序.将名为 b 的函数应用于输入的 e .

In short, this is just function application. Apply the function named b to the input e.

为了娱乐和学习而重写

我很欣赏您可以使用函数来模仿结构-这是常见的编程语言类分配.您可以将一个 Bag a 和另一个 Bag a 然后合并在一起，例如通过添加两个单独袋子的计数来返回新的 countB -很酷.

I appreciate that you can use functions to imitate structures - it's a common programming language class assignment. You can take a Bag a and another Bag a then union them, such as returning a new countB by adding the counts of the two individual bags - cool.

...但这似乎太多了.在继续进行作业之前(我想是吗?)，您可能应该对基础知识有所了解.

... but this seems too much. Before moving on with your assignment (did I guess that right?) you should probably become slightly more comfortable with the basics.

如果重写不带类型别名的函数，可能会更容易:

It might be easier if you rewrite the functions without the type alias:

emptyB :: a -> Int
emptyB = \e -> 0
-- or: emptyB e = 0
-- or: emptyB _ = 0
-- or: emptyB = const 0

袋子或没有袋子，这只是一个功能.

Bag or no bag, it's just a function.

countB :: Eq a => (a -> Int) -> a -> Int
countB b e = b e

采用 a 并生成 Int 的函数可以被赋予一个值(变量 e 的类型为 a )，并生成一个 Int .

A function that takes an a and produces an Int can... be given a value (the variable e is of type a) and produce an Int.

这篇关于不了解此haskell代码中的内容的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！