java - 没有错误!找不到java.lang.StackOverflowError的解决方案

Question

import java.util.*;

public class NRootx {

    public static Double NRmethod (Double num, int root, Double guess){
        guess= num/(root+1); //first guess  

        Double guess_output=1.0;

        for (int i=1;i<=root; i++)
        {
            guess_output *= guess;
        }

        Double error= guess_output-num;
        error = error * 0.001;
        guess = guess - error;

        if (Math.abs(error)>=0.000001)
            return NRmethod(num, root, guess); //Recursion 
        else
            return guess;

           //I can't find out the problem now. Java.lang.StackOverFlow error is showing. 

        }
    }
}

// Main function 
public static void main(String[] args) {

    Double x;
    int n;
    Scanner sc= new Scanner (System.in);
    System.out.println("Enter the value of x: ");
    x=sc.nextDouble();
    System.out.println("Enter the value of n: ");
    n=sc.nextInt();
    Double guess=x/(n+1);
    Double ans= NRmethod(x,n,guess);
    System.out.println("The value of y is ="+ans);
}

请有人帮我解决这个问题。我厌倦了这样做很多次。

Answer 1

您永远不会更改root和num，并且在每次调用guess= num/(root+1)时重置guess );，所以你会得到无休止的递归，导致 StackOverflowError。

我不确定正确的逻辑应该是什么，但为了防止无限递归，您必须将不同的 root 值或 num 值传递给每个递归调用，或者不要在方法开始时重置 guess。