Roots of cubic using C and GSL

Question

I am trying to write a C program using GSL to find the roots of a cubic equation following the instructions here: http://www.gnu.org/software/gsl/manual/html_node/Cubic-Equations.html. This is what I came up with:

#include <stdio.h>
#include <gsl/gsl_poly.h>

double *x0,*x1,*x2;
int roots;

int
main (void)
{
    roots = gsl_poly_solve_cubic(0,0,0,x0,x1,x2);

    printf( " %d ", roots);

    return 0;
}

The arguments are 0,0,0 because I wanted to test if it works first. The code compiles but when run, it crashes with no output.

What am I doing wrong?

score 2 · Accepted Answer · answered Sep 16 '12 at 17:34

2

x0, x1 and x2 are just dangling pointers - change the code to:

double x0,x1,x2;
int roots;

int
main (void)
{
    roots = gsl_poly_solve_cubic(0,0,0,&x0,&x1,&x2);

    printf( " %d ", roots);

    return 0;
}

answered Sep 16 '12 at 17:34

Paul R

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Oh right...gosh that was silly! Thanks, will accept in 10 mins. :) – Legendre Sep 16 '12 at 17:36

score 2 · Answer 2 · edited May 23 '17 at 12:21

2

You're misunderstanding how reference semantics are implemented in C. Please read this answer I just wrote on the exact same topic.

Solution:

double x0, x1, x2;

int roots = gsl_poly_solve_cubic(0, 0, 0, &x0, &x1, &x2);

Nutshell: The caller must take the address-of the recipient variable. The recipient variable must exist.

edited May 23 '17 at 12:21

Community

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answered Sep 16 '12 at 17:35

Kerrek SB

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score 1 · Answer 3 · answered Sep 16 '12 at 17:35

According to you lonk we have gsl_poly_solve_cubic (double a, double b, double c, double * x0, double * x1, double * x2). You declare 3 double pointers without allocating any memory ... this will result in a segfault.
Try to declare double variables and just pass their address :

#include <stdio.h>
#include <gsl/gsl_poly.h>


double x0,x1,x2;
int roots;

int
main (void)
{
    roots = gsl_poly_solve_cubic(0,0,0,&x0,&x1,&x2);

    printf( " %d ", roots);

    return 0;
}

Roots of cubic using C and GSL

3 Answers3