On Saturday, February 19, 2022, Stefan Pochmann <smpochm...@gmail.com>
wrote:
> >>> 1e200.is_integer()
> True
>
> So that could losslessly be converted to int, and then the division would
> succeed
If the float argument isn't an integer, you can multiply both sides by a
power of 2 that makes it one (if it's finite, obviously), so this would
still work.
Currently, int/int and float/float are always properly rounded. It would be
nice if you could just say that a/b is properly rounded whenever a and b
are int or float, and forget about conversion issues. So I'm +1 on this.
It would make certain divisions more expensive than they currently are, but
only those where one argument is float and the other is an int that isn't
exactly representable as a float (meaning, in particular, its absolute
value is greater than 2**53). It seems unlikely to me that anyone is doing
a lot of divisions of that kind and needs the speed, but if they are and
do, casting to float beforehand is an easy workaround.
Also, 10**400/float('inf') should return +0.0 instead of raising
OverflowError, and 10**400/float('nan') should be NaN. The division should
behave as though both arguments are promoted to arbitrary-precision
floating point (but the result is rounded to 64-bit float). I wouldn't
suggest that if it was hard to implement, but I think it's pretty easy
when long_true_divide already exists.
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