At 05:59 PM 11-30-2000 +0000, Nicholas Clark wrote:
>On Thu, Nov 30, 2000 at 12:46:26PM -0500, Dan Sugalski wrote:
(Note, Dan was writing about "$a=1.2; $b=3; $c = $a + $b")
>$a=1; $b =3; $c = $a + $b
>
>
> > If they don't exist already, then something like:
> >
> > newscalar a, num, 1.2
> > newscalar b, int, 3
> > newscalar c, num, 0
> > add t3, a, b
>
>and $c ends up a num?
>why that line "newscalar c, num, 0" ?
>It looks to me like add needs to be polymorphic and work out the best
>compromise for the type of scalar to create based on the integer/num/
>complex/oddball types of its two operands.
I think the "add t3, a, b" was a typo, and should be "add c, a, b"
Another way of looking at it, assuming that the Perl6 interpreter is
stack-based, not register-based, is that the sequence would get converted
into something like this:
push num 1.3 ;; literal can be precomputed at compile time
dup
newscaler a ;; get value from top of stack
push int 3 ;; literal can be precomputed at compile time
dup
newscaler b
push a
push b
add
newscaler c
The "add" op would, in C code, do something like:
void add() {
P6Scaler *addend;
P6Scaler *adder;
addend = pop(); adder = pop();
push addend->vtable->add(addend, adder);
}
it would be up to the addend->vtable->add() to figure out how to do the
actual addition, and what type to return.
> > But that probably doesn't help much. Let me throw together something more
> > detailed and we'll see where we go from there.
>
>Hopefully it will cover the above case too.
>
>Nicholas Clark
