C 数组/指针语义：如何使结构内的数组具有相应指针的属性/语义？

Question

Consider this code:

void f_01( int a, char** flags )
{
    printf("flags %p\n", flags);
    if ( ! flags ) 
    {
        return; // early exit
    }
    //some code
}

struct s01
{
    int     a;
    char**  flags1;     /* NULL = not used  */
    char*   flags[];    /* NULL = not used  */
} s_01 = { 13 };

int main()
{
    struct s01* s = &s_01;
    f_01( 1, NULL );
    f_01( 2, s->flags1 );//for demonstration purpose
    f_01( 3, s->flags ); //s->flags is expected to be NULL, but instead
                         //it evaluates to address of s->flags (i.e. &s->flags)
    return 0;
}

Actual output:

f_01 1 (nil)
f_01 2 (nil)
f_01 3 0x601038

Expected output (imaginary):

f_01 1 (nil)
f_01 2 (nil)
f_01 3 (nil)

Question: how to make an array struct member ( flags in the example) to have properties/semantics of corresponding pointer ( char** in this example), ie to be initialized to NULL by default?

The purpose of this: to be able to use the same logic in the API's implementation ( f_01 in this example) with char** argument for both NULL pointers and empty arrays . Ie inside the API there is NO additional if (! *flags) check because it is expected that if flags != NULL then it is non-empty array . It is logical, right? Then why need to add additional if (! *flags) check to handle empty arrays ? Can an empty array be evaluated somehow to NULL pointer in C?

Answer 1

The reason why the struct doesn't initialize the flags as NULL is because the last flags is "flexible array member" and not a pointer. The struct is a way of allocating enough space for all of the things that you want to put in it. So, when the struct has int a; , char** flags1; , char* flags[]; . It needs to allocate a chunk of memory that is at least sizeof(a) (probably 16 bits), sizeof(char**) (probaly 32 bits), and sizeof(char*[]) (in this particular case it is unknown). Because we don't know the size of the array of pointers yet, it will treat it as if it is zero bits and you will need to allocate enough additional bits for the size of the array when you figure out the size you need. (see http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1570.pdf §6.7.2.1 18-21)

So in this case it initializes the char* pointer as NULL because it wasn't given a value and assumes you want an array of char* that is zero in size because it wasn't given a value for its size. This array simply starts at the end of of the struct (in your case 0x601038 ). So when you pass the array into f_01 you are giving it the starting location of the zero array.

I know that you probably don't want to hear this but the only solution that I can think of is to replace that char* flags[]; with char** flags . Maybe you could make char* flags[]; change into int sizeOfFlags; and char* flagData[]; then just feed sizeOfFlags into f_01. because NULL == 0 if sizeOfFlags == 0 it will still give you NULL when the size is zero. However that is definitely the wrong way to do it.