Re: typecheck code

[Dave Hylands]

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
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