>>>>> "Camm" == Camm Maguire <camm@enhanced.com> writes:
Camm> Greetings! I think the problem is here (opers.lisp):
Camm> (defmfun mul2* (x y)
Camm> (cond
Camm> #+cl ((and (numberp x) (numberp y)) (f* x y))
Camm> ((=1 x) (simplifya y nil))
Camm> ((=1 y) (simplifya x nil))
Camm> (t (simplifya `((mtimes) ,x ,y) nil))))
Camm> You probably mean (f* x y) -> (* x y) or numberp -> fixnump. This
Camm> code compiles to the following obviously incorrect C:
Nice work. I think the best solution would be (* x y). Why it wants
f* is strange. I wish I knew the difference between the intent of mul
and mul*. Something about simplified args or results?
Ray