On Mit, 2004-06-30 at 12:01, Jan Hudec wrote: > 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 Probably you should since it is defined there. > > what gcc does. if ((type *)0) was not located at address 0, this macro > > would not work. Then the compiler is broken or not a C compiler. And that *is* the point. All your "0" and "NULL" above are the C source representation independent of the bitmap used on the hardware at run-time. And it must always 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. The pointer subtraction is only defined if both pointers point to the same array. Yes, I know that this is (almost) never checked and works as above (at least with the gcc). > all addresses were moved by a constant, it would still work. Bernd -- Firmix Software GmbH http://www.firmix.at/ mobil: +43 664 4416156 fax: +43 1 7890849-55 Embedded Linux Development and Services -- Kernelnewbies: Help each other learn about the Linux kernel. Archive: http://mail.nl.linux.org/kernelnewbies/ FAQ: http://kernelnewbies.org/faq/
