Allocating memory for multiplying two matrices

Question

A few while back I wrote this code to multiply two matrices. The function mat_mult is called in the main method. I wanted to write something which could take unknown matrix sizes and then allocate memory as needed. I seem to be getting a problem when I output the resultant matrix in the main method. If I do a printf in the mat_mult method the matrix entries are just fine.

Why is the C matrix in the main method not showing correctly? Also the way I am freeing the memory, is it the right method?

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

void mat_mult(int* p, int* q, int* r, int numP)
{
    int i, j, k, sumC=0;

    for (i = 0; i < numP; i++)
    {
        for (j = 0; j < numP; j++)
        {
            for (k = 0; k < numP; k++)
            {
                sumC = sumC + p[i*numP + k]*q[k*numP + j];
            }
            r[i*numP + j] = sumC;
            sumC = 0;
        }
    }
    for(i = 0; i < 3; i++)
    {
        for (j = 0; j < 3; j++)
        {
            printf("%d ", r[i*numP + j]);
        }
        printf("\n");
    }
}

int main()
{
    int A[3][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
    int B[3][3] = {{3, 2, 1}, {6, 5, 4}, {9, 8, 7}};


    int i, j;
    int sizA = sizeof(A)/sizeof(A[0]);

    //int C[3][3] = {0};

    /* sizA is the width of the array */
    int **C = calloc(sizA, sizeof(int*));
    for(i = 0; i < sizA; i++)
    {
        /* sizA is the height */
        C[i] = calloc(sizA, sizeof(int));
    }

    for(i = 0; i < 3; i++)
    {
        for (j = 0; j < 3; j++)
        {
            printf("%d ", C[i][j]);
        }
        printf("\n");
    }
    mat_mult(A[0], B[0], C[0], sizA);


    for(i = 0; i < 3; i++)
    {
        for (j = 0; j < 3; j++)
        {
            printf("%d ", A[i][j]);
        }
        printf("\t");
        for (j = 0; j < 3; j++)
        {
            printf("%d ", B[i][j]);
        }
        printf("\t");
        for (j = 0; j < 3; j++)
        {
            printf("%d ", C[i][j]);
        }
        printf("\n");
    }
    free(C[0]);
    free(C);
    return 0;
}

Here is the output from the program above.

0 0 0
0 0 0
0 0 0
42 36 30
96 81 66
150 126 102
1 2 3   3 2 1   42 36 30
4 5 6   6 5 4   150 126 102
7 8 9   9 8 7   0 0 0

Process returned -1073741819 (0xC0000005)   execution time : 12.472 s
Press any key to continue.

PROBLEM RESOLVED

@Jayesh

Thank you very much friend. Your last suggestion worked. I am only reproducing relevant parts of the code.

    int i, j;
    int sizA = sizeof(A)/sizeof(A[0]);

//    int C[3][3] = {0};

    /* sizA is the width of the array */
//    int **C = calloc(sizA, sizeof(int*));
//    for(i = 0; i < sizA; i++)
//    {
//        /* sizA is the height */
//        C[i] = calloc(sizA, sizeof(int));
//    }

    int (*C)[sizA] =  calloc(sizA*sizA, sizeof(int*));
    /* Rest of the Code*/

    free(C);

And here is how the output looks like.

0 0 0
0 0 0
0 0 0
42 36 30
96 81 66
150 126 102
1 2 3   3 2 1   42 36 30
4 5 6   6 5 4   96 81 66
7 8 9   9 8 7   150 126 102

Process returned 0 (0x0)   execution time : 0.085 s
Press any key to continue.

Still wondering why the first method did not work. But thanks again to all of you and Jayesh in particular.

Answer 1

Also the way I am freeing the memory, is it the right method?

No( call free about how many times you called calloc\\malloc ). To free the memory use

for(i = 0; i < sizA; i++)
{
 free(C[i]); 
}
free(C);

I prefer method for memory allocation of matrix like

int (*C)[sizA] =  calloc(sizA*sizA, sizeof(int*));

And just free it by

free(C);

See more details about different ways to allocate array memory

Answer 2

The problem is in the way that you're allocating memory and using it. Within your matmult function you have this code:

r[i*numP + j] = sumC;

which treats the result matrix r as a single-dimensioned linear array. That's fine and will work, but it means that you must also allocate it that way. Your main routine then becomes this:

int main()
{
    int A[3][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
    int B[3][3] = {{3, 2, 1}, {6, 5, 4}, {9, 8, 7}};

    int i, j;
    int sizA = sizeof(A)/sizeof(A[0]);

    /* sizA is the width of the array */
    int *C = calloc(sizA*sizA, sizeof(int));

    for(i = 0; i < 3; i++)
    {
        for (j = 0; j < 3; j++)
        {
            printf("%d ", C[i*3+j]);
        }
        printf("\n");
    }
    mat_mult(A[0], B[0], C, sizA);

    for(i = 0; i < 3; i++)
    {
        for (j = 0; j < 3; j++)
        {
            printf("%d ", A[i][j]);
        }
        printf("\t");
        for (j = 0; j < 3; j++)
        {
            printf("%d ", B[i][j]);
        }
        printf("\t");
        for (j = 0; j < 3; j++)
        {
            printf("%d ", C[i*3+j]);
        }
        printf("\n");
    }
    free(C);
    return 0;
}

An alternative approach would be to address all three arrays in this way rather than using multidimensional indexing. It's not pretty but it works.

Alternatively, for a nicer syntax, you might consider C++ for this kind of application.