c++ - 帮助解决堆栈溢出问题!

Question

首先，我想说这是我第一次尝试为我的编程课做一个递归函数!无论如何，分配是使用递归(绝对没有迭代)找到任何正整数的根。我的代码正确计算了任何正数的平方根，但是当我尝试计算一个数的四次方根或三次方根时，出现堆栈溢出错误。我将发布我的代码，我们将不胜感激任何帮助。如果您觉得需要，请开火哈。

#include<iostream>
#include<iomanip>
#include<cstdlib>
using namespace std; 

double squareRoot (int root, double number, double low, double high); 

int main() {
    int root = 0; 
    double number; 
    double low, high; 
    double guess = 0;
    double error = 0; 
    cout.setf(ios::fixed);
    cout.setf(ios::showpoint); 

    do
    {
        cout << "Find what root? (0 ends): "; 
        cin >> root; 
        if (root == 0)
        {
            exit(1); 
        }
        else
        {
            cout << "Root of what value? "; 
            cin >> number;
        }
        low = number - (number - 1);  
        high = number;
        cout << "root " << root << " of " << setprecision(4) << number << " is " << setprecision(10) << squareRoot (root, number, low, high) << endl; 
    }while (root != 0);  

    cin.get(); 
    cin.get(); 
    return 0; 
}

double squareRoot (int root, double number, double low, double high)
{
    cout.setf(ios::fixed); 
    cout.setf(ios::showpoint); 
    cout.precision(10); 
    double guess = (high + low) / double(root); 
    double error = abs(number - (guess * guess));  
    if ((guess * guess) == number || error <= 0.000000000001)
    {   
        return guess; 
    }
    else if ((guess * guess) > number)
    {
        high = guess;  
        return squareRoot(root, number, low, high); 
    }
    else
    {
        low = guess; 
        return squareRoot(root, number, low, high); 
    }
}

Answer 1

你会得到一个堆栈溢出，因为你在无限递归；你永远找不到答案。

拿笔和纸，通过一个输入(比如 3 代表根，8 代表值)来完成你的递归......弄清楚为什么你的解决逻辑不起作用。