Address of array - difference between having an ampersand and no ampersand

Question

I have a struct that looks like this:

struct packet {
  int a;
  char data[500];
};
typedef struct packet packet_t;

I\'m a little confused

Answer 1

I don't know why this was voted down, it's a good question that exposes a confusing behaviour of C.

The confusion comes because normally when you define an array a real pointer is created:

char data[100];
printf("%p\n", data);    // print a pointer to the first element of data[]
printf("%p\n", &data);   // print a pointer to a pointer to the first element of data[]

So on a typical 32 bit desktop system 4 bytes are allocated for data, which is a pointer to 100 chars. Data, the pointer, itself exists somewhere in memory.

When you create an array in a struct no pointer is allocated. Instead the compiler converts references to packet.data into a pointer at runtime but does not allocate any memory for storing it. Rather it just uses &packet + offsetof(data).

Personally I would prefer the syntax to be consistent and require an ampersand, with packet.data generating some kind of compile time error.