[Steven D'Aprano <st...@pearwood.info>[\
> Alas, math.remainder goes through float:
>
> >>> math.remainder(3**500, 3) # Should be 0.
> 1.0
> >>> math.remainder(3**500 + 2, 3) # Should be 2.
> 1.0
You mean -1 here: remainder() returns a member of the equivalence
class with least absolute value, and abs(-1) < abs(2).
> It would be nice if remainder() worked properly for exact values,
> without overflow or rounding errors.
I don't think so, It's required by 754-like standards explicitly for
floating-point values. In that way, it's much too like math.fmod:
>>> import math
>>> math.fmod(10**300, 100)
60.0
That's intended. And, of course, so is this:
>>> math.remainder(10**300, 100)
-40.0
It's far more important that remainder() work similarly to fmod() than
that remainder() implement some bigint function nobody wants ;-)
Seriously, modulus for floating point is overwhelmingly used as part
of argument-reduction steps. Indeed, almost the entire actual point
to it is that
0 <= remainder(x, y) <= abs(y) / 2.0
so that it leaves behind "a remainder" of magnitude as small as
possible, which in floating point algorithms can greatly speed
convergence of following approximations.
> ...
> I don't know whether this counts as compelling or not, but I have a
> divmod variant which returns an always positive modulus for use in base
> conversion. It is only used in one place in a module I haven't touched
> in a decade, so it is quite possible that if I were writing that code
> again today, I wouldn't do it that way.
Insufficient information to guess from here. If; you're mucking around
with negative bases, no, I expect that's too exceedingly niche to
count as compelling. But if you're doing % with a positive base, the %
result Python returns right now is already always non-negative.
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