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Re: [Scheme-reports] [scheme-reports-wg1] John Cowan's votes and rationales on the seventh ballot
On 09/19/2012 02:49 AM, Alex Shinn wrote:
> On Wed, Sep 19, 2012 at 3:39 PM, Mark H Weaver<mhw@x> wrote:
>> 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.
> That definition only applies when z is a number.
Says who? I don't see why it shouldn't apply whenever 'z' is something
that can be multiplied, and +nan.0 can be multiplied. Furthermore,
contrary its name, (number? +nan.0) => #true. In fact (real? +nan.0) =>
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