C++ 平方根函数错误

C++ Square Root Function Bug

我有一个计算整数平方根的 C++ 算法。该程序可以正常工作,但有一个缺陷。它无法计算小于 1 的数字的平方根。例如,它无法计算 .5 或 .9 或 .0000001 等的平方根,但在所有其他情况下都按计划工作。我设置了 X,所以它不允许负输入,但我仍然不明白为什么它不会 return 小于 1 的值。

include <iostream>
#include <cmath>
#include <cassert>

using namespace std;

double squareroot(double x)

{ /* computes the square root of x */

  /* make sure x is not negative .. no math crimes allowed! */
    assert(x >= 0);
    if (x == 0) return 0;

    /* the sqrt must be between xhi and xlo */
    double xhi = x;
    double xlo = 0;
    double guess = x / 2;

    /* We stop when guess*guess-x is very small */

    while (abs(guess*guess - x) > 0.000001)
    {
        if (guess*guess > x)  xhi = guess;
        else xlo = guess;
        guess = (xhi + xlo) / 2;
    }

    return guess;
}

/* Test Stub */


int main()
{
    double testvalue;
    cout << "\n Enter a TestValue= ";
    cin >> testvalue;
    cout << endl;
    double testresult = squareroot(testvalue);
    cout << "\n Square Root= " << testresult << "\n";
}

感谢您的帮助!我能够使用以下方法解决问题:

if (x<1) {
    xhi = 1;
    xlo = x;
    guess = (x + 1) / 2;
}

0.5 的平方根约为 0.7。如果检查失败,您的逻辑是猜测一个较小的数字。你需要做的是添加一个额外的检测层来查看数字是否<1,然后修改流程以增加而不是减少下一个猜测。

添加一些调试输出。这将帮助您理解为什么程序在 x < 1.0.

时永远不会收敛到解。
while (abs(guess*guess - x) > 0.000001)
{
    if (guess*guess > x)
    {
       cout << "Changing xhi\n";
       xhi = guess;
    }
    else
    {
       cout << "Changing xlo\n";
       xlo = guess;
    }

    guess = (xhi + xlo) / 2;
    cout << "guess: " << guess << endl;
}

如果遵循the Newton-Raphson method,程序会收敛得更快。此外,无论 x 是大于还是小于 1,它都有效。

double squareroot(double x)
{
   /* make sure x is not negative .. no math crimes allowed! */
   assert(x >= 0);
   if (x == 0) return 0;

   double guess = x / 2;
   while (abs(guess*guess - x) > 0.000001)
   {
      double dx = 0.5*(guess*guess -x)/guess;
      guess -= dx;
      cout << "guess: " << guess << endl;
   }

   return guess;
}

在 x<1 的情况下,您需要更改初始边界,因为平方根不在 0 和 x 之间,而是在 x 和 1 之间

double xhi, xlo, guess;
if (x > 1){
    xhi = x;
    xlo = 0;
    guess = x / 2;
}
else{
    xhi = 1;
    xlo = x;
    guess = (x + 1) / 2;
}