Segmentation Fault, c program

Question

Why does these code returning a segmentation fault? I searched online but can't find the problem. Could you please help? Is it because of pointer malloc fault?

void multiply(int **a, int n, int m, int **b, int m2, int p, int **c)
{
    int i, j, k;
    for (i = 0; i < n; i++){
        for (j = 0; j < p; j++){
            for (k = 0; k < m; k++){
                c[i][j] += a[i][k] * b[k][j];
                }
            printf("%d", c);}}
}
main()
{
    int **a, **b, **c;
    int n, m, p;
    int i, j;
    scanf("%d", &n); scanf("%d", &m); scanf("%d", &p);
    a = (int **)malloc(n*sizeof(int*));
    for (i = 0; i < n; i++){
        a[i] = (int*)malloc(m*sizeof(int));}
    for (i = 0; i < n; i++){
        for (j = 0; j < m; j++){
            scanf("%d", &a[i][j]);}
    }
    b = (int **)malloc(m*sizeof(int*));
    for (i = 0; i < m; i++){
        b[i] = (int *)malloc(p*sizeof(int));}
    for (i = 0; i < m; i++){
        for (j = 0; j < p; j++){
            scanf("%d", &b[i][j]);}
    }
    c = (int **)malloc(n, sizeof(int));
    for (i = 0; i<n; i++)
        c[i] = (int *)malloc(p, sizeof(int));
    multiply(a, n, m, b, m, p, c);
    free(a);
    free(b);
    free(c);
}

Thanks^^

I edit some code and it is now correct. Thank you guys so much. Cheers!

Is there some reason why you can't just debug this in the usual way, e.g. step through the code in your debugger and see what's going on ? — Paul R, Mar 03 '15 at 17:10
multiply(**a, n, m, **b, m, p, **c); -> multiply(a, n, m, b, m, p, c); i can't imagine a compiler not warning about that. — ryanpattison, Mar 03 '15 at 17:13
Enable you're compiler warnings to pedantic levels. There are multiple issues in your code they will identify, and *all* of them need to be fixed. — WhozCraig, Mar 03 '15 at 17:14
free(a), free(b), free(c); you called malloc and friends more than 3 times.. — ryanpattison, Mar 03 '15 at 17:17
multiply(**a, n, m, **b, m, p, **c);->I'm sorry that it's a fault of copying... — facebook-331114790430340, Mar 03 '15 at 17:35

score 1 · Answer 1 · answered Mar 03 '15 at 17:12

1

If you're on an LP64 architecture (int is 32 bits and int* is 64 bits), c is too small, you should use

c = (int **)calloc(n, sizeof(int*));

answered Mar 03 '15 at 17:12

JeremyP

84,577
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score 1 · Answer 2 · edited May 23 '17 at 12:28

1

You have a typo here:

        for (k = 0; k < m; j++){
                           ^

This should be:

        for (k = 0; k < m; k++){
                           ^

Also:

multiply(**a, n, m, **b, m, p, **c);

should be:

multiply(a, n, m, b, m, p, c);

Note that if you had turned on compiler warnings then the compiler would have helpfully pointed out this mistake to you.

Note also that you should not cast the result of malloc in C.

edited May 23 '17 at 12:28

Community

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1

answered Mar 03 '15 at 17:12

Paul R

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Look at the function call ..multiply() that is the issue and hence segmentation fault – Gopi Mar 03 '15 at 17:13
I think there is more than one issue, but this one certainly needs to be fixed. – Paul R Mar 03 '15 at 17:13

Segmentation Fault, c program

2 Answers2