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Re: [Scheme-reports] [scheme-reports-wg1] John Cowan's votes and rationales on the seventh ballot



On 09/19/2012 01:58 AM, John Cowan wrote:
> Mark H Weaver scripsit:
>
>> On the contrary, (expt<anything>  0) should yield an exact 1
>
> Are you sure?  What about (expt +nan.0 0)?

Yes, (expt +nan.0 0) => 1.  When the exponent is an exact non-negative 
integer, then (expt z k) may be defined as (* z z z z ...) with 'k' 
occurrences of 'z'.  Therefore (expt z 0) => (*) => 1.

      Mark

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