Remainder ( % ) operator and GCC

[Frans Englich]

On Wednesday 09 November 2005 19:38, corey taylor wrote:
> You could use fmod.

But isn't fmod for floating point mod computation? From what I've understood 
"%" when invoked with integers beforms integer remainder compuation, as 
opposed to floating point.


Cheers,

		Frans


>
> corey
>
> On 11/9/05, Frans Englich <frans.englich@telia.com> wrote:
> > On Wednesday 09 November 2005 19:19, Ian Lance Taylor wrote:
> > > Ladislav Mecir <lmecir@mbox.vol.cz> writes:
> > > > Could you give me an advice how to make the following program (and
> > > > programs using many times the % operator) run reliably when compiled
> > > > by GCC on different operating systems even when the user gives -1 as
> > > > the B value?
> > > >
> > > > #include <stdio.h>
> > > > int main(int argc, char *argv[]) {
> > > >     int a = -2147483647;
> > > >     int b;
> > > >     printf ("The value of B: ");
> > > >     scanf ("%i", &b);
> > > >     printf ("%i %% %i: %i", a, b, (a - 1) % b);
> > > >     return 0;
> > > > }
> > >
> > > Don't use '%'.  It is machine dependent when used with a negative
> > > number.
> >
> > What should one use instead? What is the portable, machine-independent
> > alternative?
> >
> >
> > Cheers,
> >
> >                 Frans