OpenEuphoria

Re: remainder() is not right

• Posted by CChris <christian.cuvier at agriculture.g??v.fr> May 05, 2008
• 1075 views

Euler German wrote:
> 
> > On 4 May 2008 at 16:40, CChris wrote (maybe snipped):
> 
> > It returns *a* emainder. When a and b are positive, this is the
> > expected value. When signs are otherwise, then a discrepancy appears
> > with what, in arithmetic, is called "remainder".
> > 
> 
> Please, I'm not discussing mathematical discrepancies. I only stated 
> that remainder(), as described in reference manual, works as said, 
> thus returning the "left over" of a division, so it can't be told 
> wrong. I'm NOT saying there's no need for a "signed_remainder()", 
> though I have no use for it. YMMV.
> 
> Some quick explanation at:
> - <a
> href="http://en.wikipedia.org/wiki/Remainder">http://en.wikipedia.org/wiki/Remainder</a>
> - <a
> href="http://en.wikipedia.org/wiki/Modulo_operation">http://en.wikipedia.org/wiki/Modulo_operation</a>
> 
> IMO you're describing a modulo function (or operator) so, if we had 
> something like:
> 
> m = mod(x, y) this would be in Euphoria as:
> 
> }}}
<eucode>
> m = x - y * floor(x / y)
> </eucode>
{{{

> 
> ...and produces those same results you claim.
> 
> Best,
> Euler
> 
> PS: This message was sent earlier but didn't find a safe way to the 
> list. Maybe munged headers and body. Sorry if you're getting this 
> twice.
> 
> -- 
> _
> _| euler f german
> _| sete lagoas, mg, brazil
> _| efgerman{AT}gmail{DOT}com
> _| -----------------------------
> _| Reply preferably to the list,
> _| or to the address above. Thx!
> 
> 

Ok for mod() or modulo(), and the formula would be right. Since that's what I
mostly need, I have learned not to use remainder() but when both operands are
above 0 (with extension to sequences).

CChris