Re: Zeller's Algorithm

[Robert A Duff <bobduff@shell01.TheWorld.com>]

Hyman Rosen <hyrosen@mail.com> writes:

> Darren New wrote:
> > Robert A Duff wrote:
> >>Doesn't this make the % operator totally useless for negative numbers?
> > Yes.
> 
> No. It is required in C or C++ that for non-zero b,
> 	a / b * b + a % b == a
> 
> When division rounds towards zero, as is usually desired
> (the C++ standard claims that this is what the Fortran
> standard requires), this forces the remainder operation
> to return negative results. Eg., we want -3 / 2 == -1,
> so we must have -3 % 2 == -1 for the requirement to hold.

Hmm.  You're just "blaming" the fact that "%" is ill-defined on the fact
that "/" is ill-defined.  Wouldn't it be easier to understand that the
code is portable (and correct), if rounding were deterministic?
Can you give an example of an algorithm that takes advantage of the
above equivalence, but doesn't care which way rounding goes?

It's OK to blame this language flaw on history (C was implemented before
being standardized, unlike Ada -- we could have a nice thread discussing
the advantages and disadvantages of each approach), but to claim it's
not a flaw seems wrong.

By the way, how often do folks use "mod" or "rem" on negative numbers?
In *my* code, the numbers are (almost?) always nonnegative (and the
compiler knows that, because of subtype ranges, and can optimize
accordingly).

If I need to know the difference between "mod" and "rem", I have to look
it up every time, because I can't remember it.

- Bob