求解二次；学习函数；程序不返回值

Question

请看一下我在第七章C编程之后作为赋值的一部分写的程序，该程序应该根据用户键入的常量的值返回二次方的根。这个程序应该很简单，我是初学者。尽管编译器确实编译了程序，但我得到的唯一输出是提示消息。但是在我输入了三个值之后，什么也没有发生，程序结束，我回到了我的终端。我已经编辑了这个程序。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:

因此，main function语句仍然出现。

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:

你看到这个格式了吗？这一切为什么要发生？

#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返回扫描成功的参数个数。在本例中，您希望它是3，而不是1。

Answer 2

这条线

return x1, x2;

不返回两个值。这一行相当于：

x1;  // Nothing happens.
return x2;

如果您希望能够返回两个值，请创建一个struct并返回struct的一个实例。

typedef struct pair { float x1; float x2;} pair;

// Change the return type of solve_quadratic 
pair solve_quadratic (float a, float b, float c);

当您准备好从solve_quadratic返回时，使用：

pair p;
p.x1 = x1;
p.x2 = x2;
return p;

并更改您使用它的位置：

// if (scanf ("%f %f %f", &a, &b, &c) == 1)
//                                      ^^^ That is not correct.
if (scanf ("%f %f %f", &a, &b, &c) == 3)
{
    pair p = solve_quadratic(a,b,c);
    printf("Square roots of this quadratic are: %f %f \n", p.x1. p.x2);
}