c - 求解二次；学习功能；程序不返回值

Question

请看一下我在第 7 章 C 编程之后作为作业的一部分编写的程序。该程序应根据用户输入的常量值返回二次的根。该程序应该非常简单；我处于初学者水平。虽然编译器确实编译了程序，但我得到的唯一输出是提示信息。但是在我输入三个值之后，没有任何反应，程序结束，我又回到了我的终端。我已经编辑了程序。现在，如果我输入一些使判别式小于 0 的值，我会得到

Please, enter three constants of the quadratic:   
1 2 3
Roots are imaginary
Square roots of this quadratic are:

这样main函数语句还是出现了。

如果我输入其他值，我会得到

Please, enter three constants of the quadratic:   
1 8 2
-0.256970 and -7.743030Square roots of this quadratic are:

你看到这种格式了吗？为什么会这样？

#include <stdio.h>
#include <math.h>

float abs_value (float x);
float approx_sqrt (float x);
float solve_quadratic (float a, float b, float c);

// The main function prompts the user for 3 constant values to fill a quadratic
// ax^2 + bx + c
int main(void)
{
    float a, b, c;

    printf("Please, enter three constants of the quadratic:   \n");

    if (scanf ("%f %f %f", &a, &b, &c) == 3)
        printf("Square roots of this quadratic are:  \n", solve_quadratic(a,b,c));

    return 0;
}

// Function to take an absolute value of x that is used further in square root function
float abs_value (float x)
{
    if (x < 0)
        x = - x;
    return x;
}

// Function to compute an approximate square root - Newton Raphson method
float approx_sqrt (float x)
{
    float guess = 1;

    while (abs_value (pow(guess,2) / x) > 1.001 || abs_value (pow(guess,2) / x) < 0.999)
          guess = (x / guess + guess) / 2.0;

    return guess;
}

// Function to find roots of a quadratic
float solve_quadratic (float a, float b, float c)
{
    float x1, x2;
    float discriminant = pow(b,2) - 4 * a * c;

    if (discriminant < 0)
    {
        printf("Roots are imaginary\n");
        return -1;
    }
    else
    {
        x1 = (-b + approx_sqrt(discriminant)) / (2 * a);
        x2 = (-b - approx_sqrt(discriminant)) / (2 * a);
    }

    return x1, x2;
}

谢谢!

Answer 1

if (scanf ("%f %f %f", &a, &b, &c) == 1)

scanf 返回成功扫描的参数数量。在这种情况下，您希望它是 3，而不是 1。