了解返回反向计数的递归循环 [英] Understanding recursive loop that returns an inverted count

问题描述

inner conditional: 5
inner conditional: 4
inner conditional: 3
inner conditional: 2
inner conditional: 1
0

创建上述输出的函数如下，我完全理解它是如何工作的.

The function that created the above output is below and I perfectly understand how it works.

function loop(counter){

    if(counter > 0){
        console.log("inner conditional: " + counter );
        return loop(counter - 1)
    }

    console.log(counter);
    return counter
}

loop(5)

<小时>

当修改函数使得内部条件不返回函数时，console.log 会显示以下内容.

---

When the function is modified so that the inner condition does not return the function the console.log displays the following.

inner conditional: 5
inner conditional: 4
inner conditional: 3
inner conditional: 2
inner conditional: 1
0
1
2
3
4
5

下面是负责上述输出的函数.

Below is the function that is responsible for the above output.

function loop(counter){

    if(counter > 0){
        console.log("inner conditional: " + counter );
        loop(counter - 1)
    }

    console.log(counter);
    return counter
}

loop(5)

我完全不明白这一点.我只想解释一下创建扩展前向计数发生了什么.

I do not understand this at all. I would simply like an explanation as to what is happening to create the extended forward count.

我使用以下工具来可视化逐步执行代码，但我仍然没有理解"它:http://pythontutor.com/javascript.html#mode=display

I used the following tool to visualize stepping through the code and I still don't "get" it: http://pythontutor.com/javascript.html#mode=display

推荐答案

我一直在寻找的答案是 1-5 计数正在从堆栈顶部拉出.

The answer that I was looking for is that the 1-5 count is being pulled off the top of the stack.

我发现了一个很棒的 YouTube 视频，它在此处通过可视化解释了这一点:

I found a great youtube video that explains this here with visualizations:

https://www.youtube.com/watch?v=s8JpA5MjYac

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