递归函数找出数的幂 [英] Recursion function to find power of number

问题描述

我正在写一个递归函数来找到一个数字的强度，它似乎是编译但不输出任何东西。

I'm writing a recursion function to find the power of a number and it seems to be compiling but doesn't output anything.

#include <iostream>

using namespace std;

int stepem(int n, int k);

int main()
{
    int x, y;

    cin >> x >> y;

    cout << stepem(x, y) << endl;

    return 0;
}

int stepem(int n, int k)
{
    if (n == 0)
        return 1;
    else if (n == 1)
        return 1;
    else
        return n * stepem(n, k-1);
}

我试过调试它，它说的问题是这行：
return n * stepem（n，k-1）;

I tried debugging it, and it says the problem is on this line : return n * stepem(n, k-1);

k似乎有些奇怪值，但我不知道为什么？

k seems to be getting some weird values, but I can't figure out why?

推荐答案

你应该检查指数k，变化。

You should be checking the exponent k, not the number itself which never changes.

int rPow(int n, int k) {
    if (k <= 0) return 1;
    return n * rPow(n, --k);
}

你的k值越来越奇怪，因为你会一直计算，直到你用完内存基本上，你会创建许多堆栈帧，k去-infinity（假设）。

Your k is getting weird values because you will keep computing until you run out of memory basically, you will create many stack frames with k going to "-infinity" (hypothetically).

这就是说，理论上可能的编译器给你警告它永远不会终止 - 在这个特定的情况。但是，自然不可能解决这个一般（查找Halting问题）。

That said, it is theoretically possible for the compiler to give you a warning that it will never terminate - in this particular scenario. However, it is naturally impossible to solve this in general (look up the Halting problem).

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