On Thu, Dec 07, 2000 at 01:14:40PM +0000, David Mitchell wrote:
> Dan Sugalski <[EMAIL PROTECTED]> wrote:
> > All the math is easy if the scalars are of known types. Addition and
> > multiplication are easy if only one of the scalars involved is of known
> > type. Math with both of unknown type, or subtraction and division with the
> > right-hand operand of unknown type, is rather more difficult. :(
>
> I'm not clear with your scheme how addition works if one of the scalars
> (the adder) is of unknown type. ie
>
> given sv1 of type NUM, sv2 of type UNKNOWN; $sv1 + $sv2 would invoke:
>
> sv1->add[UNKNOWN](sv2),
>
> which somewhere will cause a function in the vtable for NUMs to be called, eg
>
> NUM_add_UNKNOWN(sv1,sv2) { .... }
> Now, how does does this function perform its calculation?
I'm guessing that Dan is planning to take advantage of addition and
multiplication being commutative.
sv1->add[UNKNOWN](sv2)
swaps to
sv2->add[NUM](sv1)
(It's "obvious" in the usual way - not obvious until you see it. I've been
prodding in pp_add in perl5, so I've been thinking about these sort of
things)
Nicholas Clark
