Re: Zero-size array as struct member

Joe Greer <>
Wed, 18 Aug 2010 19:18:25 +0000 (UTC)
Vladimir Jovic <> wrote in news:i4gueb$vqv$1

If the array's size is zero, how can you access even one element?

What's your point? Why would you do that? If b were a vector, you would
get similar behavior. That is, it is just as undefined if you don't
bother to check the size of the vector before assigning data. This is a
idiom used by device drivers to return information requiring a minimum
number of allocations and a minimum number of frees. It is always used
with malloc and the size of the array is always embedded in the struct
somewhere so that you don't go out of bounds. So, to fix up your code

The example similar to the original example, except this one compiles :

#include <iostream>
using namespace std;
struct A
     int a;

       int b_len;

     int b[0];

A * AFactory(int sz)
   A * p = (A *)malloc(sizeof(A) + sz * sizeof(b[0]));
   p->b_len = sz;
   return p

int main()
     A *p = new A;

       A *p = AFactory(0);

     p->a = 5;

     // We know we didn't allocate space so we wouldn't bother with
this, but...
       if (p->b_len >= 3)

     p->b[0] = 6;
     p->b[1] = 7;
     p->b[2] = 8;


     std::cout << "p->a="<<p->a << std::endl;

       for (int ix = 0; ix < p->b_len; ++ix)
     std::cout << "p->b[" << ix << "] = " << p->b[ix] << std::endl;

     free(p); // you could also have function for this, but
deallocation doesn't require additional info.


I would never advocate the above as a C++ application pattern, but if you
are messing in drivers and kernel, then every clock cycle counts and
multiple heap hits can kill performance.

This is one of the places I would really like to see some sort of VLA
available (at least if it allocated memory in this fashion). The nice
thing is that it is one heap allocation and free. Rolling your own as
above is a bit of a pain. Fortunately, I haven't had to do stuff like
that in years.


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