Passing multidimensional arrays to a function

#include <assert.h>
#include <stdlib.h>

/* When passing a multidimensional array (i.e. an array of arrays) to a
   function, it decays into a pointer to the first element as usual.  But only
   the top level decays, so what is passed is a pointer to an array of some fixed
   size (4 in this case). */
void f(int x[][4]) {
    assert(sizeof(*x) == sizeof(int) * 4);
}

/* This prototype is equivalent to f(int x[][4]).
   The parentheses around *x are required because [index] has a higher
   precedence than *expr, thus int *x[4] would normally be equivalent to int
   *(x[4]), i.e. an array of 4 pointers to int.  But if it's declared as a
   function parameter, it decays into a pointer and becomes int **x, 
   which is not compatable with x[2][4]. */
void g(int (*x)[4]) {
    assert(sizeof(*x) == sizeof(int) * 4);
}

/* An array of pointers may be passed to this, since it'll decay into a pointer
   to pointer, but an array of arrays may not. */
void h(int **x) {
    assert(sizeof(*x) == sizeof(int*));
}

int main(void) {
    int foo[2][4];
    f(foo);
    g(foo);

    /* Here we're dynamically creating an array of pointers.  Note that the 
       size of each dimension is not part of the datatype, and so the type 
       system just treats it as a pointer to pointer, not a pointer to array
       or array of arrays. */
    int **bar = malloc(sizeof(*bar) * 2);
    assert(bar);
    for (size_t i = 0; i < 2; i++) {
        bar[i] = malloc(sizeof(*bar[i]) * 4);
        assert(bar[i]);
    }

    h(bar);
    
    for (size_t i = 0; i < 2; i++) {
        free(bar[i]);
    }
    free(bar);
}