用于计算平方根的自定义程序未显示任何输出 [英] Custom program to calculate sqare root isn't showing any output

问题描述

这是一个用户定义的程序，用于打印数字的平方根.它也应该适用于也不是完美平方的数字.每次将 i 递增 0.01 的步长，并检查 i * i 的值是否等于 n .如果相等，则打印 i 的值.

This is a user defined program to print the square root of a number. It is supposed to work for numbers which aren't perfect squares as well. i is incremented by a step of 0.01 each time and the the value of i*i is checked if equal to n. And if equal then the value of i is printed.

#include <stdio.h>

void squareRoot(double);

int main()
{
    double num;
    scanf("%lf", &num);
    squareRoot(num);
    return 0;
}

void squareRoot(double n)
{
    double i;
    for (i = 0; i < n; i += 0.01)
    {
        //printf("%.2lf\n",i*i);
        if (i * i == n)
        {
            printf("%lf\n", i);
            break;
        }
    }
}

推荐答案

由于注释中所说明的原因，您使用的方法充其量是非常不准确的，对于该示例我有几种计算平方根的方法将使用 巴比伦方法 使用简单的算术运算非常容易实现:

The method you use is, at best, very inaccurate for the reasons explained in the comments, there are several ways to calculate a square root, for this sample I'll use the Babylonian method which in its simplified form is very easy to implement using trivial arithmetic operations:

> 运行示例

#include <stdio.h>

double squareRoot(double);

int main() {

    double num;

    printf("Enter number: \n");
    scanf("%lf", &num);

    if(num < 0) {
        puts("Negative values not allowed");
        return 1;
    }

    printf("Square root of %.2lf is %lf", num, squareRoot(num));
}

double squareRoot(double num) {

    double sqroot = num / 2, temp = 0;

    while (sqroot != temp) {
        temp = sqroot;
        sqroot = (num / temp + temp) / 2;
    }
    return sqroot;
}

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