From jmc at r-project.org  Sat Oct 25 21:55:21 2008
From: jmc at r-project.org (John Chambers)
Date: Sat, 25 Oct 2008 15:55:21 -0400
Subject: [Rd] 0 ^ NaN == Inf, why?
In-Reply-To: <490376DE.101@r-project.org>
References: <54BCF12B6BE6423DBC6CA23B9B7D5D23@dell16>
	<490376DE.101@r-project.org>
Message-ID: <490379A9.3090903@r-project.org>

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