Multidimensional arrays follow the same rules as single-dimensional arrays when passing them to a function. However the combination of decay-to-pointer, operator precedence, and the two different ways to declare a multidimensional array (array of arrays vs array of pointers) may make the declaration of such functions nonintuitive. The following example shows the correct ways to pass multidimensional arrays.
#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);
}
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