The paragraph and example of eqv? on procedures with local state are superfluous: it doesn't matter if a procedure has local state or not, as the report has explicitly un-specified eqv? on procedures. In this sense, closures are not objects.
In 6.2.4, "A NaN always compares false to any number, including a NaN". I thought the result of the discussion was that (eqv? +nan.0 +nan.0) could be #t? I didn't really follow it anyway. Andy -- http://wingolog.org/ _______________________________________________ Scheme-reports mailing list [email protected] http://lists.scheme-reports.org/cgi-bin/mailman/listinfo/scheme-reports
