Thread: Free()

When I compile this code on gcc using linux ubuntu 11.04

Code:

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

float *mult_matrix(int n, float *p, float *q);
float *get_matrix(int m, int n);
void show_matrix(int m, int n, float *p);

int main(void)
{
    int n;
    float *a, *b, *x;

    printf("\nEnter the size of the linear system: \n");
    scanf("%d", &n);

    a=get_matrix(n, n);
    show_matrix(n, n, a);
    b=get_matrix(n, 1);
    show_matrix(n, 1, b);
    x=mult_matrix(n, a, b);
    show_matrix(n, 1, x);

    free(a);
    free(b);
    free(x);

    return 0;
}

float *mult_matrix(int n, float *p, float *q)
{
    int i, j;
    float *x;

    x=(float *)malloc((n)*sizeof(float));    

    if (!x) {
        printf("NULL pointer detected. Exiting ...");
        exit(1);
    }

    for(i=0; i<n; i++) {
        *x=0;
        x++;
    }
    x=x-n;

    for (i=0; i<n; i++) {
        for(j=0; j<n; j++) {
            *x=*x+(*p)*(*q);
            p++;
            q++;
            x++;
        }    
        q=q-n;
    }
    x=x-n;

    return x;
}

float *get_matrix(int m, int n)
{
    int i, j;
    float *a;
    
    a=(float *)malloc((m*n)*sizeof(float));

    if (!a) {
        printf("NULL pointer detected. Exiting ...");
        exit(1);
    }

    for (i=0; i<m; i++) {
        for (j=0; j<n; j++) {
            printf("\nEnter with the element a%d%d of the matrix:", i+1, j+1);
            scanf("%f", a);
            a++;
        }    
    }

    a=a-m*n;

    return a;
}

void show_matrix(int m, int n, float *p)
{
    int i, j;
    
    printf("\n");
    for (i=0; i<m; i++) {
    printf("\n");
        for (j=0; j<n; j++) {
            printf(" %f ", *p);
            p++;
        }
    }
    printf("\n");
    p=p-m*n;

}

I got this message on the end of the program after running it normally on the terminal:

Code:

*** glibc detected *** ./a.out: double free or corruption (out): 0x097d5058 ***

It means that the program or the compiler frees the pointers automatically so I dont have to use free()? Or this dont make any sense at all? Please, I need some help. Cheers.

No, the compiler never "frees" anything for you. You must always do that yourself.

What that error usually means is
1. You are trying to free something that has already been freed.
2. You are trying to free something that has never been allocated and the pointer contains a garbage address

You probably have an instance of case 2, since you only free in one place, and it's not in a loop, it's hard to free something twice. You most likely overwrite the value in a, b or x at some point so you no longer have the right address to pass to free.

What input are you giving it? What output do you expect and what output do you actually get? I get a seg fault, not a double free/corruption error. It's probably related to the same bug, but I can't be sure without your input so I can try to reproduce exactly what you're doing.

Also, have you run this through a debugger? Do you know how? Try compiling like so:

Code:

$ gcc -Wall -g matrix.c

Then, load up your program in gdb

Code:

$ gdb ./a.out

Run it, and keep entering input until it crashes. Then, get a backtrace to show where the crash happened

Code:

(gdb) run
...
Program received signal SIGSEGV, Segmentation fault.
0x080487d6 in mult_matrix (n=3, p=0xb7eedfe8, q=0xb7eefff4) at matrix.c:50
50                  *x=*x+(*p)*(*q);
(gdb) backtrace
#0  0x080487d6 in mult_matrix (n=3, p=0xb7eedfe8, q=0xb7eefff4) at matrix.c:50
#1  0x08048712 in main () at matrix.c:20

Use up, down and print commands to move around the stack frame and examine variables to find out what went wrong.

Originally Posted by guitarman

I got this message on the end of the program after running it normally on the terminal:

Code:

*** glibc detected *** ./a.out: double free or corruption (out): 0x097d5058 ***

It means that the program or the compiler frees the pointers automatically so I dont have to use free()? Or this dont make any sense at all? Please, I need some help. Cheers.

It means that you are freeing something that was never allocated, presumably because you perform pointer arithmetic directly on the pointer you allocate and get it wrong.

You could try this for example.

Code:

float *get_matrix(int m, int n)
{
    int i, j;
    float *a;
     
    a=(float *)malloc((m*n)*sizeof(float));

printf("%p\n", a); // <--

    if (!a) {
        printf("NULL pointer detected. Exiting ...");
        exit(1);
    }
 
    for (i=0; i<m; i++) {
        for (j=0; j<n; j++) {
            printf("\nEnter with the element a%d%d of the matrix:", i+1, j+1);
            scanf("%f", a);
            a++;
        }    
    }
 
    a=a-m*n;

printf("%p\n", a); // <--

    return a;
}

Well, one problem I see:
How many floats do you allocate for x? Yep, n. Now, you have nested for loops on lines 48-56. The body of the inner loop is executed n * n times, and each time you do x++ without resetting x. Thus, you start accessing memory you don't own. That's undefined behavior and a big problem.

You do this in a strange manner. I would recommend allocating 2-d arrays so you can just use array indexing instead of your wonky (and error-prone) pointer arithmetic ju-ju. Read this: Question 6.16

Originally Posted by anduril462

Well, one problem I see:
How many floats do you allocate for x? Yep, n. Now, you have nested for loops on lines 48-56. The body of the inner loop is executed n * n times, and each time you do x++ without resetting x. Thus, you start accessing memory you don't own. That's undefined behavior and a big problem.

You do this in a strange manner. I would recommend allocating 2-d arrays so you can just use array indexing instead of your wonky (and error-prone) pointer arithmetic ju-ju. Read this: Question 6.16

Thanks for your help. I made some modifications on function mult_matrix() and the problem seems to be solved

Code:

float *mult_matrix(int n, float *p, float *q)
{
    int i, j;
    float *x;

    x=(float *)malloc((n)*sizeof(float));    

    if (!x) {
        printf("NULL pointer detected. Exiting ...");
        exit(1);
    }

    for(i=0; i<n; i++) {
        *x=0;
        x++;
    }
    x=x-n;

    for (i=0; i<n; i++) {
        for(j=0; j<n; j++) {
            *x=*x+(*p)*(*q);
            p++;
            q++;
            x++;
        }    
        q=q-n;
        x=x-n; // new code
    }

    return x;
}

I'll take a look anyway on the link you posted. Cheers.