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

[Mark H Weaver <mhw@x>]

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) => 
#true.

      Mark

_______________________________________________
Scheme-reports mailing list
Scheme-reports@x
http://lists.scheme-reports.org/cgi-bin/mailman/listinfo/scheme-reports