Hi Sri,
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
>
> I appreciate any explain.
If dummy and dummy2 are of different types, then when you try and do a
pointer comparison (&dummy == &dummy2) it will produce a compiler
warning/error.
The actual comparison will always fail, but it doesn't matter since
the results aren't used.
typecheck always returns 1.
Dave Hylands
_______________________________________________
Kernelnewbies mailing list
Kernelnewbies@xxxxxxxxxxxxxxxxx
http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
