递归在c ++因子程序 [英] Recursion in c++ Factorial Program

问题描述

hello我有一段代码，我基于一些其他递归和阶乘程序
编码，但我的问题是，我真的很困惑，如何存储的值，保留它，然后返回它在end

hello i have this piece of code that i coded based on some other recursion and factorial programs but my problem is that i am really confused as to how it stored the value and kept it and then returned it at the end

int factorialfinder(int x)
{
    if (x == 1)
    {
        return 1;
    }else
    {
        return x*factorialfinder(x-1);
    }
}
int main()
{
  cout << factorialfinder(5) << endl;
}

所以5进去，再次调用它的函数，乘以4一次又一次，然后它得到一个，它返回阶乘答案

so 5 goes in, and gets multiplied by 4 by calling its function again and again and again, then it gets to one and it returns the factorial answer

为什么？我不知道它是如何存储的，为什么返回1返回实际的答案，它真正做什么？

why? i have no idea how it got stored, why is return 1 returning the actual answer, what is it really doing?

推荐答案

img src =http://i.stack.imgur.com/q2Zkv.gifalt =来自IBM开发人员网站的递归图片>

资料来源：Image取自： IBM开发人员网站

Source: Image is taken from: IBM Developers website

只要看看上面的图片，你会明白更好。

Just take a look at the picture above, you will understand it better. The number never gets stored, but gets called recursively to calculate the output.

所以当你调用事实（4）时，当前堆栈用于存储每个参数为递归调用发生在factorialfinder（1）。所以计算如下：5 * 4 * 3 * 2 * 1。

So when you call the fact(4) the current stack is used to store every parameter as the recursive calls occur down to factorialfinder(1). So the calculation goes like this: 5*4*3*2*1.

int factorialfinder(int x)         
{
    if (x == 1)        // HERE 5 is not equal to 1 so goes to else
    {
        return 1;
    }else
    {
        return x*factorialfinder(x-1); // returns 5*4*3*2*1  when x==1 it returns 1
    }
}

希望这有帮助。

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