Re: [Caml-list] behaviour of mod

[Stephen Dolan <stephen.dolan@cl.cam.ac.uk>]

On Sun, Jan 18, 2015 at 6:48 PM, Hannes Mehnert <hannes@mehnert.org> wrote:
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> Hi,
>
> is the behaviour of modulo arithmetics intentional:
>  -5 mod 4 = -1 ?
>
> While this reflects the C behaviour, my expectation was to always have
> a positive result:
>  -5 mod 4 = 3
>
> Any hints?

Given OCaml's integer division operator, this is the correct "mod".

The important property is:

    (x / y) * y + (x mod y) = x

In other words, (x mod y) gives the error by which (x / y) * y fails to equal x.

There are two reasonable (/, mod) pairs which have this behaviour. The
first, which C and OCaml use, is where (x / y) rounds towards zero and
(x mod y) has the same sign as x, so -5 / 4 = -1 and -5 mod 4 = -1.
The second is where (x / y) rounds down and (x mod y) has the same
sign as y, so -5 / 4 = -2 and -5 mod 4 = 3.

Subjectively, I think the first division operator (round-toward-zero,
aka truncate) and the second modulo operator are the more natural. The
second modulo operator actually implements modular arithmetic, since
with it x mod n buckets the integers x into n equivalence classes
differing by multiples of n. But using the first (/) and the second
mod breaks the remainder property above.

Incidentally, Haskell provides both: the first is called (quot, rem)
while the second is (div, mod).

Stephen