计算平方根时的 SegFault 错误(牛顿法)

2022-01-12 00:00:00 segmentation-fault c++

我对 C++ 很陌生，但我知道有很多方法可以接收 SegFault 错误，但我不确定我为什么会在这里遇到一个.该程序的目标是使用牛顿法计算数字的平方根.我假设它与递归有关，但我很确定代码会在 Java 中运行.

I'm very new to C++, but I'm aware that there are tons of ways to receive a SegFault error, but I'm not sure why I'm getting one here. The goal of the program is to compute the square root of a number using Newton's method. I'm assuming it has something to do with the recursion, but I'm pretty sure that the code would run in Java.

#include <iostream>
#include <sstream>

using namespace std;

double sqrt(double eps, double num, double last_guess)
{

    if(num == 0 || num == 1)
    {
        return num;
    }

    int next = (num + (num/last_guess)) / 2;

    if(abs(last_guess - next) <= eps)
        return next;
    else
        return sqrt(eps, num, next);
}

int main(int argc, char *argv[]) {
    double eps, num;
    istringstream iss; //input string

    if(argc == 3)
    {
        iss.str(argv[1]);
        if ( !(iss >> eps) ) {
            cerr << "Error: The first argument is not a valid double." << endl;
            return 1;
        }

        iss.clear();

        iss.str(argv[2]);
        if ( !(iss >> num) ) {
            cerr << "Error: The second argument is not a valid double." << endl;
            return 1;
        }

    } else if(argc == 2) {
        iss.str(argv[1]);
        if ( !(iss >> num) ) {
            cerr << "Error: Argument is not a valid double." << endl;
            return 1;
        }

        eps = 0.000001;
    }

    cout << "sqrt(" << eps << ", " << num << ") = " << sqrt(eps,num, num) << endl;
    return 0;
}

样本输入: 0.000001 4.0

Sample input: 0.000001 4.0

推荐答案

在 GDB 下运行你的程序显示它处于无限递归中:

Running your program under GDB shows that it is in infinite recursion:

(gdb) run
Starting program: /tmp/a.out 0.000001 4.0

Program received signal SIGSEGV, Segmentation fault.
0x000055555555522d in sqrt (eps=<error reading variable: Cannot access memory at address 0x7fffff7feff8>, 
    num=<error reading variable: Cannot access memory at address 0x7fffff7feff0>, last_guess=<error reading variable: Cannot access memory at address 0x7fffff7fefe8>) at foo.cc:7
7       {
(gdb) bt 20
#0  0x000055555555522d in sqrt (eps=<error reading variable: Cannot access memory at address 0x7fffff7feff8>, 
    num=<error reading variable: Cannot access memory at address 0x7fffff7feff0>, last_guess=<error reading variable: Cannot access memory at address 0x7fffff7fefe8>) at foo.cc:7
#1  0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=3) at foo.cc:19
#2  0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=2) at foo.cc:19
#3  0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=3) at foo.cc:19
#4  0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=2) at foo.cc:19
#5  0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=3) at foo.cc:19
#6  0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=2) at foo.cc:19
#7  0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=3) at foo.cc:19
#8  0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=2) at foo.cc:19
#9  0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=3) at foo.cc:19
#10 0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=2) at foo.cc:19
#11 0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=3) at foo.cc:19
#12 0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=2) at foo.cc:19
#13 0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=3) at foo.cc:19
#14 0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=2) at foo.cc:19
#15 0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=3) at foo.cc:19
#16 0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=2) at foo.cc:19
#17 0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=3) at foo.cc:19
#18 0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=2) at foo.cc:19
#19 0x00005555555552df in sqrt (eps=9.9999999999999995e-07, num=4, last_guess=3) at foo.cc:19
(More stack frames follow...)

  1. 没有理由让你的例程递归.

  1. There is no reason to make your routine recursive.

您的算法无法识别(缺少检查)您已经计算出正确答案.

Your algorithm fails to recognize (is missing a check) that you've already computed the correct answer.

您不应将猜测之间的增量与 epsilon 进行比较.您应该比较计算出的答案和真实答案之间的差异.

You shouldn't compare the delta between your guesses to the epsilon. You should compare the delta between your computed answer and the real answer instead.

正如@PaulMcKenzie 所说，您不应该将逐次逼近值存储在整数中(使用 double 代替).

As @PaulMcKenzie noted, you shouldn't store your successive approximations in a integer (use double instead).

要修正程序，你需要使用正确的公式进行下一次猜测:

To correct the program, you need to use correct formula for the next guess:

    double next = (last_guess + (num/last_guess)) / 2;

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