(Moved over to -internals, since it's not really a parser API thing)
At 11:06 AM 11/30/00 -0600, Jarkko Hietaniemi wrote:
>Presumably. But why are you then still talking about "the IV slot in
>a scalar"...? I'm slow today. Show me how
>
> $a = 1.2; $b = 3; $c = $a + $b;
>
>is going to work, what kind of opcodes do you see being used?
>(for the purposes of this exercise, you may not assume the optimizer
> doing $c = (1.2+3) behind the curtains :-)
Okay, assuming $a, $b, and $c alread exist, in pseudo-opcodes:
newscalar t1, num, 1.2
newscalar t2, int, 3
scalar_assign a, t1
scalar_assign b, t2
newscalar t3, num, 0
add t3, a, b
scalar_assign c, t3
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
But that probably doesn't help much. Let me throw together something more
detailed and we'll see where we go from there.
Dan
--------------------------------------"it's like this"-------------------
Dan Sugalski even samurai
[EMAIL PROTECTED] have teddy bears and even
teddy bears get drunk
