• simendsjo (2/2) Jul 01 2011 What is contained within this byte?
• Ali =?iso-8859-1?q?=C7ehreli?= (21/23) Jul 01 2011 void* can point to any data, in which case it is considered to be
• simendsjo (2/25) Jul 05 2011 Needed some time to digest your answer, but it makes sense now. Thanks.

simendsjo <simendsjo gmail.com> writes:

What is contained within this byte?
(T[0]).sizeof == 0, why isn't void also 0?

Ali =?iso-8859-1?q?=C7ehreli?= <acehreli yahoo.com> writes:

On Fri, 01 Jul 2011 21:18:45 +0200, simendsjo wrote:

 What is contained within this byte?
 (T[0]).sizeof == 0, why isn't void also 0?

void* can point to any data, in which case it is considered to be 
pointing at the first byte of the data. Having a size of one makes it 
point to the next byte when incremented:

    int i;
    void * v = &i;   // first byte
    ++v;             // second byte

Similarly, an empty struct has a size of one:

import std.stdio;

struct S
{}

void main()
{
    assert(S.sizeof == 1);
}

But in that case it is needed to identify S objects from one another just 
by having different addresses. The following array's data will occupy 10 
bytes:

    S[10] objects;
    assert(&(objects[0]) != &(objects[1]));

Ali

simendsjo <simendsjo gmail.com> writes:

On 01.07.2011 22:18, Ali Çehreli wrote:
 On Fri, 01 Jul 2011 21:18:45 +0200, simendsjo wrote:

 What is contained within this byte?
 (T[0]).sizeof == 0, why isn't void also 0?

 void* can point to any data, in which case it is considered to be
 pointing at the first byte of the data. Having a size of one makes it
 point to the next byte when incremented:

      int i;
      void * v =&i;   // first byte
      ++v;             // second byte

 Similarly, an empty struct has a size of one:

 import std.stdio;

 struct S
 {}

 void main()
 {
      assert(S.sizeof == 1);
 }

 But in that case it is needed to identify S objects from one another just
 by having different addresses. The following array's data will occupy 10
 bytes:

      S[10] objects;
      assert(&(objects[0]) !=&(objects[1]));

 Ali

Needed some time to digest your answer, but it makes sense now. Thanks.