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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.
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