Why does program output x1 = Nan, x2 = Nan?

Question

I can't work out why my program isn't doing what it's supposed to do. The task was to make a c program that can solve quadratic using two c files. This is the code called interface.c that we got to work with, nothing has to be changed on this one:

#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;
}

Next is the code I made which isn't working, there seems to go something wrong with the integer types I think. With every quadratic formula it will output x1 = nan x2 = nan. There is something wrong with the integer type but can't figure out which.

#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

`discriminant > 0`? Are you supposed to call the function by that name? — Some programmer dude, Mar 22 '16 at 15:37
Yeah, I suspect that, in order to avoid use of a function pointer, you need 'discriminant()'. — Martin James, Mar 22 '16 at 15:41
..something that would never even have been an issue if the coefficients were passed into discriminant() as parameters. — Martin James, Mar 22 '16 at 15:44
i found the issue. in the if statement it should have been discriminant()>0 — MrAlphaking, Mar 22 '16 at 15:53
Also fix: `(-b) / (2*a);` --> `(-b) / (2.0*a);` integer vs FP division — chux - Reinstate Monica, Mar 22 '16 at 15:54
but there still seems to go something wrong with x1imag and x2imag, the output values aren't correct — MrAlphaking, Mar 22 '16 at 16:06
"something wrong" and "output values aren't correct" is not as helpful as posting 1) inputs, 2) outputs and 3) expected outputs. (Suspect you have not taken [this comment](http://stackoverflow.com/questions/36158645/why-does-program-output-x1-nan-x2-nan#comment59955663_36158645) into account yet. — chux - Reinstate Monica, Mar 22 '16 at 16:12
i added the inputs outputs and expected output below the code — MrAlphaking, Mar 22 '16 at 16:26

nariuji · Accepted Answer · 2016-03-22T16:45:06.817

0

Quadratic formula?

conditional expression of「b^2-4ac≧0」 is not called, is it?

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;
}

(An addition)

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

edited Mar 22 '16 at 16:45

answered Mar 22 '16 at 15:54

nariuji

288
1
6

See [comment](http://stackoverflow.com/questions/36158645/why-does-program-output-x1-nan-x2-nan#comment59955663_36158645) – chux - Reinstate Monica Mar 22 '16 at 15:55
Thank you for helping me out! – MrAlphaking Mar 22 '16 at 15:56
Now you are computing the discriminant three times with every call to abc. And you don't take into account the possibility that `a` is zero. Also, don't use `pow(b,2)` to compute a square - just use `b*b`. You also compute the same square root twice. – FredK Mar 22 '16 at 16:13

Why does program output x1 = Nan, x2 = Nan?

1 Answers1