Meaning of char arr[0]

Question

Possible Duplicate:
Array of zero length

I have seen such type of code :-

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

struct pkt {
  int pk_type;
  int tag;
  int magic_num;
  char data[0];  <-------- what does this mean ???
}__attribute__((packed));

typedef struct pkt p_t;
int main(void)
{
  p_t *p = (p_t *)calloc(1, sizeof(p_t));
  if(!p) {
    perror("p");
    return -1;
  }
  p->pk_type = 1;
  p->tag = 0xb00bbabe;
  p->magic_num = 0xcafebabe;
  strcpy("hello world", p->data); <-- Seg faults here !!!  

  return 0;

}

I was wondering the meaning of arr[0]. I mean in what scenarios do we need to use arr[0]. What purpose does it serve?

Answer 1

In C, the members of a struct are always allocated in the order they appear. So, my_pkt->data is a pointer "one past the end" of a pkt object. If initialized with

my_pkt = malloc( sizeof( struct pkt ) + 50 );

then my_pkt->data points to the beginning of a 50-byte general-purpose buffer.

Only the final member of a struct is allowed to be defined this way.

To be more compliant with C99, omit the 0 and write char data[];.

Answer 2

It's a flexible array member, that can be used, when you need a structure of varying size. It simplifies the allocation of the extra size, just see the link.

Answer 3

This is what the GCC Manual has to say about zero length arrays

Answer 4

This is a zero byte array. Used to append data of arbitrary size at runtime. Very useful in sending data over a socket.

You should refer this for more details- http://bytes.com/topic/c/answers/547407-char-data-0-a

Answer 5

char arr[0] is a same as char *. And the reason you are getting segmentation fault is,because when you do calloc in following line of code

p_t *p = (p_t *)calloc(1, sizeof(p_t));

it actually initialize every memory with zero.So when you passed a null pointer,it caused segmentation fault because of null pointer.