[Rd] 0 ^ NaN == Inf, why?

[John Chambers]

A small PS:

John Chambers wrote:
>
> Along the line, notice that both R_pow and pow give 0^0 as 1.  (Just at 
> a guess, C might give 0^-0 as Inf, but I don't know how to test that in R.)
>   
I tried a little harder, and apparently the guess is wrong.  It seems 
that pow(0, -0) is 1 in C.   Would seem better to either have pow(0,0) 
and pow(0,-0) both be NaN or else 1 and Inf, but ...
> John
>
>
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