In a message dated Tue, 24 Sep 2002, Jonathan Scott Duff writes:
> On Tue, Sep 24, 2002 at 11:14:04AM -0400, Aaron Sherman wrote:
> > Again, we're wading into the waters of over-simplification. Let's try:
> >
> > sub foo1(){ my @foo=(1,2,3); return @foo; }
> > sub foo2(){ my $foo = [1,2,3]; return $foo; }
> > sub foo3(*@list) { print @list.length, "\n"; }
> > @foo = (1,2,3);
> > foo3(@foo, [1,2,3], foo2(), foo1());
> >
> > Ok, so what is the output? 12? 10? 8?
> >
> > More importantly, why? I could argue the case for each of the above
> > numbers, but I think 12 is the way it would be right now.
>
> Hrm. I think it must be 8. Since foo3() flattens it's parameters, we
> get this:
>
> foo3(1, 2, 3, [1,2,3], [1,2,3], 1, 2, 3);
>
> and since the two [1,2,3] are scalar things, we have 8 scalar things
> in our list. Splat doesn't "look inside" the thing it flattens AFAIK,
> so it doesn't flatten the two [1,2,3].
Yes, but would the lack of parens in
foo3 @foo, [1,2,3], foo2, foo1;
change anything? (I think and hope not.)
Trey
