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On Sun, Apr 13, 2014 at 3:25 PM, Tim Hill <drtimhill@gmail.com> wrote:You COULD, but it actually wouldn't make any difference. Treating it
>
> If you assume the usual mathematical identity: x^-n == 1 / (x^n), is there any reason why 2^-3 cannot be treated as 1 // (2^3) ? This keeps the “type only” rule for how the operator is interpreted (which i agree with 100%). it also avoids surprises since, like other binary operators, if both values are integers you get an integer result (with the exception of division).
>
> —Tim
as 1 // (x^n) would result in always returning 0 anyway except when
x==0 (returning NaN), x==1 (returning 1 for all n), or x ~= 0 && n==0
(returning 1 for all x).
Now, if you change that definition to be / instead of //, thus
implying that it returns float instead of int, then that's perfectly
reasonable in my book.
/s/ Adam