Re: integer exponentiation with negative exponent
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- Subject: Re: integer exponentiation with negative exponent
- From: Tim Hill <drtimhill@...>
- Date: 2014年4月14日 00:29:28 -0700
On Apr 13, 2014, at 6:07 PM, Tim Hill <drtimhill@gmail.com> wrote:
>
> 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
>
Joking at my stupidity aside .. (hush) .. it really comes down to two choices:
1 / (x^(-n) … invert the sign and do floating point division
-or-
error
As Roberto said, work1 did the former, work2 the latter. Either is a special case in a certain sense. It’s just picking the latter of two evils.
—Tim