On Wed, Sep 19, 2012 at 3:39 PM, Mark H Weaver <[email protected]> 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. -- Alex _______________________________________________ Scheme-reports mailing list [email protected] http://lists.scheme-reports.org/cgi-bin/mailman/listinfo/scheme-reports
