Question

我无法理解为什么我的程序没有做它应该做的事情。任务是制作一个可以使用两个c文件解决二次方的c程序。这是我们必须使用的名为interface.c的代码，在这一个上不需要改变：

#include <stdio.h>

void abc (void);

int a, b, c;

extern double x1real, x1imag, x2real, x2imag;

static void get_parameters (void)
{
    scanf("%d", &a);
    scanf("%d", &b);
    scanf("%d", &c);
}

void print_solution(void)
{
    printf("The roots of %dx^2 + %dx + %d are:\n",a,b,c);

    if(x1imag == 0 && x2imag == 0)
    {
        if(x1real == x2real)
        {
            printf("x = %.4f\n", x1real);
        }
        else
        {
            printf("x1 = %.4f, x2 = %.4f\n", x1real, x2real);
        }
    }
    else
    {
        printf("x1 = %.4f+%.4fi, x2 = %.4f-%.4fi\n", x1real, x1imag, x2real, x2imag);
    }
}

int main (void)
{
    int runs, run;

    scanf("%d",&runs);

    for(run=0; run < runs; run++)
    {
        get_parameters();
        abc();
        print_solution();
    }

    return 0;
}

接下来是我制作的代码不起作用，我认为整数类型似乎有问题。对于每个二次公式，它将输出x1 = nan x2 = nan。整数类型有问题，但无法确定哪个。

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

extern int a, b, c;

double x1real, x1imag, x2real, x2imag;

int discriminant(void)
{
    int discriminant;

    discriminant = (pow(b,2)-4*a*c);

    return discriminant;
}

void abc (void)
{
    if (discriminant() > 0)
    {
        x1real = (-b - sqrt(discriminant()))/(2*a);
        x2real = (-b + sqrt(discriminant()))/(2*a);
        x1imag = 0;
        x2imag = 0;
    }
    else
    {
        x1real = x2real = (-b) / (2*a);
        x1imag = (-b - sqrt(-discriminant())) / (2*a);
        x2imag = (-b + sqrt(-discriminant())) / (2*a);
    }

    return;
} 
input:
4
2   0   0
1   3   2
3   4   9
1   0   1



output:
The roots of 2x^2 + 0x + 0 are:
x = 0.0000
The roots of 1x^2 + 3x + 2 are:
x1 = -1.0000, x2 = -2.0000
The roots of 3x^2 + 4x + 9 are:
x1 = -0.6667+-2.2653i, x2 = -0.6667-0.9319i
The roots of 1x^2 + 0x + 1 are:
x1 = 0.0000+-1.0000i, x2 = 0.0000-1.0000i

suspected output:
The roots of 2x^2 + 0x + 0 are:
x = 0.0000
The roots of 1x^2 + 3x + 2 are:
x1 = -1.0000, x2 = -2.0000
The roots of 3x^2 + 4x + 9 are:
x1 = -0.6667+1.5986i, x2 = -0.6667-1.5986i
The roots of 1x^2 + 0x + 1 are:
x1 = 0.0000+1.0000i, x2 = 0.0000-1.0000i

Answer 1

二次公式？

没有调用「b ^ 2-4ac≧0」的条件表达式，是吗？

int discriminant(void)
{
    int discriminant;

    discriminant = (pow(b,2)-4*a*c);

    return discriminant;
}

void abc (void)
{

//  if (discriminant > 0)    // it doesn't call [int discriminant(void)]
    if (discriminant() >= 0)
    {
        x1real = (-b - sqrt(discriminant()))/(2*a);
        x2real = (-b + sqrt(discriminant()))/(2*a);
        x1imag = 0;
        x2imag = 0;
    }
    else
    {
        x1real = x2real = (-b) / (2*a);
        x1imag = (-b - sqrt(-discriminant())) / (2*a);
        x2imag = (-b + sqrt(-discriminant())) / (2*a);
    }

    return;
}

（补充）

static void get_parameters (void)
{
    do {
        scanf("%d", &a);
        scanf("%d", &b);
        scanf("%d", &c);
    } while(a == 0)
}

为什么程序输出x1 = Nan，x2 = Nan？

1 个答案: