------- Comment #16 from paul dot richard dot thomas at cea dot fr 2006-10-31
16:14 -------
Subject: RE: Intrinsic MOD incorrect for large arg1/arg2 and slow.
FX,
>
> ------- Comment #15 from fxcoudert at gcc dot gnu dot org
> 2006-10-31 16:05 -------
> (In reply to comment #14)
> > It also does MODULO correctly
>
> Why not use remainder{f,,l}? Is it incorrect?
I understood that remainder (a, b) = a - round (a/b) * b, whereas
mod (a, b) = a - int (a/b) * b
and modulo (a, b) = a - floor (a/b) * b
but I am ready to stand corrected on the first. The latter two are taken from
the fortran standard.
Paul
--
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=24518
