Thanks for all explanation. It really helped to understand.
Sri
On Mon, Jan 31, 2011 at 1:03 PM, Manish Katiyar <mkatiyar@xxxxxxxxx> wrote:
> On Mon, Jan 31, 2011 at 9:03 AM, Sri Ram Vemulpali
> <sri.ram.gmu06@xxxxxxxxx> 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.
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