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[Bug fortran/24518] Intrinsic MOD incorrect for large arg1/arg2 and slow.



------- 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


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