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Postings from October 2001
> > method foo ( $me, $again : $big, $boy ) {...}
> >
> > should be able to be called via either of:
> >
> > foo $obj1, $obj2 : @args;
> > or:
> > ($obj1, $obj2).foo(@args);
>
>
> This stuff brings to mind all sorts of questions:
>
> * If foo() is declared as above, what values do $me and $again have when
> the method is called as $obj.foo(@args);? Is $again undef?
I would assume it's an error. foo() expects two arguments before the
(implicit) colon, but you gave it only one.
> * What happens when foo is declared as "method foo ($a: *@_) { ... }"
> and called as "foo $a, $b: @args;"? Is $b ignored?
Error. Expected one invocant but found two.
> * Would "foo $a, $b ^: @args" work as "($a,$b)^.foo(@args)"?
No. The ^ is on the method call, not on the colon.
BTW, colon isn't an operator (it's a separator), so it can't be hyped.
> * Would (($o1,$o2),($o3,$o4))^.foo(@args); be the same as
> ($o1,$o2).foo(@args); ($o3,$o4).foo(@args); ?
Almost. It would actually be the same as:
( ($o1,$o2).foo(@args), ($o3,$o4).foo(@args) )
though the difference is trivial in a void context.
> * "method foo ($a, $b: *@_) { ... }; foo $o1, $o2: @args;" seems like a
> binding that doesn't mention :=. i.e., ($a,$b) := ($o1,$o2);
*All* argument passing to method and subroutine calls is done via
binding (even in Perl 5). Read back over my posts and you'll see that
I always express argument passing in terms like: "the first argument
is bound to the $x parameter".
^^^^^
> Will there be other "hidden" uses of the := operator?
The other implicit binding is also the same as in Perl 5.
Namely: the iteration variable (or $_) in a C<foreach>.
Damian