問(wèn)題描述

你好，我有一段基于其他遞歸和階乘程序編寫的代碼但我的問(wèn)題是我真的很困惑它是如何存儲(chǔ)值并保留它然后在最后返回它

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 進(jìn)入，通過(guò)一次又一次地調(diào)用它的函數(shù)被乘以 4，然后它得到 1 并返回階乘答案

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

為什么?我不知道它是如何存儲(chǔ)的，為什么 return 1 返回實(shí)際答案，它到底在做什么?

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

推薦答案

來(lái)源:圖片取自:IBM 開發(fā)人員網(wǎng)站

看看上面的圖片，你會(huì)更好地理解它.該數(shù)字永遠(yuǎn)不會(huì)被存儲(chǔ)，而是被遞歸調(diào)用以計(jì)算輸出.

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.

因此，當(dāng)您調(diào)用 fact(4) 時(shí)，當(dāng)前堆棧用于存儲(chǔ)每個(gè)參數(shù)，因?yàn)檫f歸調(diào)用發(fā)生到 factorialfinder(1).所以計(jì)算是這樣的: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
    }
}

希望這會(huì)有所幫助.

這篇關(guān)于C++ 階乘程序中的遞歸的文章就介紹到這了，希望我們推薦的答案對(duì)大家有所幫助，也希望大家多多支持html5模板網(wǎng)！