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.

Answer 2

Under most circumstances, an expression that has type "N-element array of T" will be converted to an expression of type "pointer to T", and its value will be the address of the first element in the array. This is what happens in the first printf call; the expression packet.data, which has type char [500], is replaced with an expression of type char *, and its value is the address of the first element, so you're effectively printing &packet.data[0].

One exception to this rule occurs when the array expression is an operand of the unary & operator; the type of the expression &packet.data is char (*)[500] (pointer to 500-element array of char).

The address of an array is the same as the address of the first element, so both calls to printf display the same value; it's just that the types of the expressions are different. To be pedantic, both expressions should be cast to void * in the printf calls (the %p conversion specifier expects a void * argument):

printf("%p\n", (void *) packet.data);
printf("%p\n", (void *) &packet.data);

Answer 3

Because it's the only reasonable thing to do besides making &packet.data cause a compile error. Integer a and the data char array are laid out sequentially in the stack, there is no dereferencing involved.

Answer 4

That's because array decays to a pointer pointing to first element in the sequence. So, packet.data address location is same as &packet.data or &packet.data[0].