When an array name is passed to a function the function?

Answer 1

All array names will implicitly convert to a pointer to the first element in the array. Note that when passing an array to a function, you must also pass the array length as a separate argument because the pointer alone cannot tell you how many elements were actually allocated to the array, let alone how many are currently in use. However, there are some exceptions. For example, a null-terminated string argument does not require a length argument as the null-terminator denotes the end of the character array. User-defined arrays can use a similar technique, using any "unused" or "invalid" value or token to denote the end of the array.

Note that the following function signatures are identical:

void f (int* a, unsigned len);

void f (int a[], unsigned len);

The latter is more readable as it makes it clear the pointer refers to an array.

When passing multi-dimensional arrays, you must add an extra level of indirection for each additional dimension:

void g (int* a[], unsigned rows, unsigned cols); // two-dimensional array

void h (int** a[], unsigned width, unsigned height, unsigned depth); // three-dimensional array

Multi-dimensional arrays can also be null-terminated or terminated by a designated token value. The canonical example of this is a global main function which accepts command line arguments. These arguments are passed through a null-terminated array of null-terminated strings (a two-dimensional array of type char).

int main (char* argv[], int argc) {

assert (argc>=1); // always at least one element

assert (argv[0] != NULL); // the first element is always the executable name (non-NULL)

assert (argv[argc-1] != NULL); // the last element is always non-NULL

assert (argv[argc] == NULL); // the one-past-the-end element is always NULL

return 0;

}