Thanks for all explanation. It really helped to understand. Sri On Mon, Jan 31, 2011 at 1:03 PM, Manish Katiyar <mkatiyar@gmail.com> wrote:
On Mon, Jan 31, 2011 at 9:03 AM, Sri Ram Vemulpali <sri.ram.gmu06@gmail.com> wrote:
Hi all,
/* * Check at compile time that something is of a particular type. * Always evaluates to 1 so you may use it easily in comparisons. */ #define typecheck(type,x) \ ({ type __dummy; \ typeof(x) __dummy2; \ (void)(&__dummy == &__dummy2); \ 1; \ })
#define typecheck_fn(type,function) \ ({ typeof(type) __tmp = function; \ (void)__tmp; \ })
Can anyone help me, explain the above code typecheck. How does (void)(&__dummy == &__dummy2) evaluates to 1
Infact I think it will never return 1, since the addresses of __dummy1 and __dummy2 have to be different (off by 4 or 8). As pointed out it is the next line that always returns 1. The purpose of this line is to throw away warnings like "Incompatible pointer comparison" or something like that (haven't tried :-)) incase there is a mismatch.
-- Thanks - Manish
-- Regards, Sri.