How to pass a 2D dynamically allocated array to a function?

See the code below. After passing the 2d array base location as a double pointer to myfunc(), you can then access any particular element in the array by index, with s[i][j].

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

void myfunc(int ** s, int row, int col) 
{
    for(int i=0; i<row; i++) {
        for(int j=0; j<col; j++)
            printf("%d ", s[i][j]);
        printf("\n");
    }
}

int main(void)
{
    int row=10, col=10;
    int ** c = (int**)malloc(sizeof(int*)*row);
    for(int i=0; i<row; i++)
        *(c+i) = (int*)malloc(sizeof(int)*col);
    for(int i=0; i<row; i++)
        for(int j=0; j<col; j++)
            c[i][j]=i*j;
    myfunc(c,row,col);
    for (i=0; i<row; i++) {
        free(c[i]);
    }
    free(c);
    return 0;
}

If your compiler supports C99 variable-length-arrays (eg. GCC) then you can declare a function like so:

int foo(int cols, int rows, int a[][cols])
{
    /* ... */
}

You would also use a pointer to a VLA type in the calling code:

int (*a)[cols] = calloc(rows, sizeof *a);
/* ... */
foo(cols, rows, a);