array as function argument

Question

Hi,

consider the following code:

#define MAX_COUNT 4
static int foo(int w[MAX_COUNT])
{
    int *p = w;

    if (w) {
        w[0] = 10; w[1] = 20; w[2] = 30; w[3] = 40;
    }

    return 0;
}

is it portable and legal pass NULL to above defined foo() (for example, for a situation when I don't need w[] be altered)? Given that the name of an array is a pointer to its first element, it seems to me that it should be OK, but perhpas I'm missing something?

Thanks !

Answer 1

It should be fine. The number of elements in the array that you passed are merely a formalism, and not really needed. w is just a pointer to int. There is no other runtime or compilation check.

Answer 2

In C, an array type function parameter is the same as a pointer type, so the following are the same:

static int foo(int w[]);
static int foo(int* w);

So, yes, it is legal to pass NULL to this function.

Given that the name of an array is a pointer to its first element

Not quite. An array decays to a pointer to its initial element under most circumstances, the exceptions being when it is the operand of sizeof or the unary & operator (the address-of operator).

Answer 3

C99 has the construct with the awfull syntax (for your example)

static int foo(int w[static MAX_COUNT]);

which basically means that the function expects at least MAX_COUNT elements. Calling such a function with NULL would then be considered as an error.

Unfortunately this feature is not yet widely implemented. e.g gcc and clang just accept the syntax but don't do anything useful with the information.