Matrix vs Malloc (2D array)

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

void testOne(int m[][2], int tam) {
    int x;
    for (x = 0; x < tam; x++) {
        m[x][0] = x;
        m[x][1] = x + 1;
    }
}

int* testTwo(int tam) {
    int* m;
    int x;
    m = malloc(sizeof(int) * 2 * tam);
    if (m == NULL) {
        printf("Error allocating memory!");
        exit(1);
    }
    for (x = 0; x < tam*2; x+=2) {
        m[x + 0] = x + 1;
        m[x + 1] = x + 2;
    }
    return m;
}

//this method is mased in something I found one Internet
//you need to go deeper, and you are doing the same as above
//http://stackoverflow.com/questions/8740195/how-do-we-allocate-a-2-d-array-using-one-malloc-statement
//the method above is what I learned at university
//I think, it's useful if you want an array (with unknown size) of pointers
int** testThree(int tam) {
    int** m, x;
    m = (int**)malloc(tam * sizeof *m + (tam * (2 * sizeof **m)));
    if (m == NULL) {
        printf("Error allocating memory!");
        exit(1);
    }
    for (x = 0; x < tam*2; x+=2) {
        m[x + 0] = (int*)1;
        m[x + 1] = (int*)2;
    }
    return m;
}

int main() {
    int a[2][2], x, d = 2;
    int* b = NULL;
    int** c = NULL;

    testOne(a, d);
    for (x = 0; x < d; x++) {
        printf("%d, %d\n", a[x][0], a[x][1]);
    }

    b = testTwo(d);
    for (x = 0; x < d*2; x+=2) {
        printf("%d, %d\n", b[x + 0], b[x + 1]);
    }
    free(b);

    c = testThree(d);
    for (x = 0; x < d*2; x+=2) {
        printf("%d, %d\n", (int)c[x + 0], (int)c[x + 1]);
    }
    free(c);

    return 1;
}