On Wed, Aug 20, 2008 at 1:30 PM, Peter Zijlstra <a.p.zijlstra@chello.nl> wrote:
quoted text
> Nick is right, try:
>
> int main(int argc, char **argv)
> {
>        unsigned int x = 7, y = 5;
>        printf("%d\n", avg(x,y));
>        return 0;
> }
>
> It fails because 5-7 = -2, which needs a signed division or sign
> extending right shift.
>
> we'd need something like:
>
> #define avg(x, y) ({                            \
>        typeof(x) _avg1 = (x);                  \
>        typeof(y) _avg2 = (y);                  \
>        (void) (&_avg1 == &_avg2);              \
>        _avg1 + (signed typeof(x))(_avg2 - _avg1)/2; })
>
> except that typeof() doesn't work that way.
>
> #define avg(x, y) ({                            \
>        typeof(x) _avg1 = (x);                  \
>        typeof(y) _avg2 = (y);                  \
>        (void) (&_avg1 == &_avg2);              \
>        _avg1 + (long)(_avg2 - _avg1)/2; })
>
> works for the above example, but when I make it long long, so as to
> match the longest supported type, it goes boom again - for as of yet
> unknown reasons.

I think you'd want to cast it with a (signed) instead? as in:

#include <stdio.h>

#define avg(x, y) ({            \
       typeof(x) _x = (x);      \
       typeof(y) _y = (y);      \
       (void) (&_x == &_y);     \
       _x + (signed)(_y - _x)/2; })

int main (void) {
        unsigned long long a=7,b=5;

        printf("%d %d\n", avg(a,b), avg(b,a));
}

...which works here, for me, but hey, I managed to goof up my other
test case, so take it for a spin.

Ray
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