On Wed, Jun 30, 2004 at 12:43:24 +0300, Nir Tzachar wrote: > On Wed, 30 Jun 2004, Jan Hudec wrote: > > > > as i've said b4, i dont see what this macro has to do with the memory > > > representation of the null pointer. > > > the ((type *)0) member is used to get a pointer to a struct 'type', > > > which is located at address 0 -> hence, the address of the member is its > > > offset in the struct. nothing to do with actual memory... > > > > No. ((type *)0) is used to get a NULL pointer of given type. The > > C specification DOES say 0 must convert to NULL and does NOT say it must > > be located on address 0. In gcc, it always is, though. > > thats my point. i dont care about C specifications, as u said before, but > what gcc does. if ((type *)0) was not located at address 0, this macro > would not work. Yes. Well, though, it could still be saved. It's the condition that: (char *)a - (char *)b == (int)a - (int)b, that's relied upon. If, eg. all addresses were moved by a constant, it would still work. ------------------------------------------------------------------------------- Jan 'Bulb' Hudec <bulb@ucw.cz>
Attachment:
signature.asc
Description: Digital signature
