Haskell:将整数推送到堆栈实现时的无限列表 [英] Haskell : Infinite list when integer is pushed to stack implementation

问题描述

我正在尝试实现一个简单的堆栈，但我很困惑为什么当我将一个整数压入堆栈时会得到一个无限列表.

I'm trying to implement a simple Stack but I'm confused as to why I get an infinite list when I push an integer to the stack.

所有其他功能都按我的预期工作，但我不明白 push 的问题.当我尝试将一个空堆栈分配给已推送如下变量的自身时，它会出错:

All of the other functions work as I expect them but I don't understand the problem with push. It goes wrong when I try to assign an empty stack to itself that has pushed a variable like the following:

λ > a = makeStack
λ > push 3 a
[3]
λ > a
[]
λ > a = push 3 a
λ > a
[3,3,3,3,3,3,3,3,3,3^CInterrupted.

type Stack a = [a]

makeStack :: Stack a 
makeStack = []

push :: a -> Stack a -> Stack a
push a as = (a:as)

推荐答案

Haskell 不允许突变.在源文件中，如果您定义一个变量 a 然后尝试重新分配它，就像您在此处使用 a = push 3 a 所做的那样，您会得到一个编译错误.您不这样做的唯一原因是您在 GHCi 中工作，它确实允许您重新定义变量 - 这纯粹是一种方便，因此您不必在尝试不同的定义时不断思考新名称.

Haskell does not allow mutation. In a source file, if you define a variable a and then attempt to reassign it, as you do here with a = push 3 a, you would get a compilation error. The only reason you don't is that you are working in GHCi, which does allow you to redefine variables - this is purely a convenience so you don't have to keep on thinking up new names while experimenting with different definitions.

而且，至关重要的是，a = push 3 a 不是基于前一个值给 a 一个新值，因为它会使用命令式语言.相反，它是a 就其自身而言的定义.

And, crucially, a = push 3 a is not giving a new value to a based on the previous one, as it would be in an imperative language. Instead, it is a definition of a in terms of itself.

这就是你得到一个无限列表的原因 - 你的定义被处理如下:

That's why you get an infinite list - your definition is processed as follows:

a = push 3 a
   = 3:a
   = 3:(push 3 a)
   = 3:(3:a)

等等.由于 Haskell 的懒惰，这样的定义没有问题 - 当您要求完整列表时，GHCi 会像这里一样一次简单地计算一个元素，因此会一直打印 3 秒，直到您告诉它停止.

and so on. Because of Haskell's laziness, there's no problem with a definition like this - GHCi will, when you ask for the full list, as here, simply calculate one element at a time, and therefore keep printing 3s until you tell it to stop.

要获得您想要的，您只需键入 push 3 a，或者如果您需要为其指定名称，只需从 a 中选择一个不同的名称.b = push 3 a 后跟 b 将按照您的预期运行.

To get what you want, you need to either just type push 3 a, or if you need to assign it a name, simply choose a different name from a. b = push 3 a followed by b will behave as you expect.

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