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.
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