That works, and also this looks good, although you can't step back more than
the mod size else it goes negative.
To step back n units, where n < RING_SIZE:
((ringpos - n + RING_SIZE) % RING_SIZE))
To prove it to myself, I wrote a test program that looped through the values,
printed them out, and I counted them all. It is attached (see next comment for
rationale).
Clearly this has been a thick-headed week for me.
Thanks for looking over my shoulder!
Aaron
Ilja Booij <[EMAIL PROTECTED]> said:
>
> Ilja Booij wrote:
> > Hi,
> >
> > the ringbuf stuff is severly broken. C module arithmetic can be pretty
> > akward:
> >
> > (ringpos - 1) % RING_SIZE, with ringpos = 0 and RING_SIZE = 6:
> >
> > -1 % 6 = -1
> >
> > WTF? why isn't it 5?
> >
> > I've just made a simple function to fix this, but I'm looking for a more
> > elegant solution.
>
> If you think compiler macros are elegant, then the elegant solution is
> here. :)
>
> I've added a macro MOD(x,y) to pipe.c which calculates the modulo of x
> with respect y. It also works when x is negative.
>
> Ilja
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>
--
#include <stdio.h>
#define RING_SIZE 6
int main() {
int ringpos = 0;
for (ringpos = 0; ringpos <= RING_SIZE; ringpos++ ) {
printf("ringpos = %d\n", ringpos);
printf( " less 1: %d\n", ((ringpos - 1 + RING_SIZE) % RING_SIZE));
printf( " less 2: %d\n", ((ringpos - 2 + RING_SIZE) % RING_SIZE));
printf( " less 3: %d\n", ((ringpos - 3 + RING_SIZE) % RING_SIZE));
printf( " less 4: %d\n", ((ringpos - 4 + RING_SIZE) % RING_SIZE));
printf( " less 5: %d\n", ((ringpos - 5 + RING_SIZE) % RING_SIZE));
printf( " less 6: %d\n", ((ringpos - 6 + RING_SIZE) % RING_SIZE));
printf( " less 7: %d\n", ((ringpos - 7 + RING_SIZE) % RING_SIZE));
printf( " less 8: %d\n", ((ringpos - 8 + RING_SIZE) % RING_SIZE));
printf( " less 9: %d\n", ((ringpos - 9 + RING_SIZE) % RING_SIZE));
}
}
