Re: Zero-size array as struct member
Joe Greer wrote:
Vladimir Jovic <vladaspams@gmail.com> wrote in news:i4gueb$vqv$1
@news.albasani.net:
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
Will next assertion pass or fail :
int a[0];
assert( sizeof( a ) == 0 );
?
What bothers me is how can you use an array of size zero.
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
below.
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
}
This looks like a hack. Why declare an array of size 0? Why not use a
pointer (like someone suggested)? Is it another (now obsolete and
depreciated) way of declaring pointers?
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