OpenEuphoria

Re: remainder() is not right

• Posted by "Euler German" <eulerg at gmail.com> May 05, 2008
• 1080 views

> On 5 May 2008 at 8:41, Matt Lewis wrote (maybe snipped):

> Euler German wrote:
> > 
> > If I got the idea right, this has to do with de *ability* to convert
> > Euphoria code into C. The modulo operator in C doesn't behave
> > exactly the same way depending of the compiler and hardware. Seems
> > they "fixed" it after ISO 1999. In a 100% Euphoria world remainder()
> > should suffice, but after translated into C would it produce the
> > same result?
> 
> This wouldn't be an issue.  If it's all euphoria, then, obviously, it
> would work the same.  If it's implemented in the backend as part of
> the runtime, it would still use that same code, rather than simply
> calling a C runtime library function.
> 
> Matt
> 

Maybe I didn't express myself clearly. What I mean is that an 
Euphoria program translated into C and compiled by a C compiler could 
produce different results depending on the C compiler and/or the 
hardware used. I was thinking on what Ricardo Forno said about it:

    <rf>
    More on the subject.
    In C, there are two ways to get an integer quotient:
    q = floor(a /(double)b);
    and
    q = (int)(a / (double)b);

    For negative "a", they give different results.
    </rf>

Hope it make sense now. :)

Best,
Euler

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