"Bob H." wrote:
> 
> "Tom Melly" <tom### [at] tomandlucouk> wrote in message
> news:3bfe2478@news.povray.org...
> > The docs refer to mod(A, B) as:
> > Formula is mod=((A/B)-int(A/B))*B

That's correct.

> >
> > Shouldn't that be:
> > Formula is mod=A-(int(A/B)*B)
> 

That's also correct.

> Not that I'd know or anything but I thought:  mod=A/B-int(A/B)

That's wrong.

> For instance, if A=3 and B=2 that would get you the modulus 0.5; 

Wrong, again.

 3 mod 2 is 1

You cannot change that fact.
 (more exactly, the fact is that 3 = 1 when considering modulo 2,
  in fact 15 = 1 too [modulo 2], and mathematicaly 15 = 3 also [...]).


>so long as
> int() rounds down anyway.

Dependant on what you call down.
int keeps only the integer part. So,it's really down for positive numbers,
but it's a kind of up for negative numbers ( int(-2.8) = -2 ).
The formula given in the doc is fine using int() as it probably is
derived from the coding.

There is usually a catch with modulo operation with negative numbers,
at least in most computer implementation. I did not test the resulting value
for:

  mod(-3,2)
  mod(3,-2)
  mod(-3,-2)

But I suspect they give some negative values as most langage do.

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