Re: integer exponentiation with negative exponent

[Tim Hill <drtimhill@...>]

On Apr 13, 2014, at 5:04 PM, Coda Highland <chighland@gmail.com> wrote:

> On Sun, Apr 13, 2014 at 3:25 PM, Tim Hill <drtimhill@gmail.com> wrote:
>> 
>> 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
> 
> You COULD, but it actually wouldn't make any difference. Treating it
> 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
> 

lol … you can tell i was tired when i wrote that can’t you?

—Tim