On Tue, Nov 22, 2005 at 09:42:38AM -0800, jerry gay wrote:
> your example in the previous message made me think. what will parrot
> do if a parrot sub declares the :outer subpragma, and the sub to which
> it refers doesn't have a lexical pad?
Nothing; that's entirely legal. And it's even useful, *if* the outer
sub has a :outer of its own.
sub foo ($a) {
my sub bar {
my sub baz { $a }
&baz; #note - this returns a reference
}
bar();
}
While &bar has no pad, it does have :outer(foo), so the find_lex in
&baz will work.
> .sub do_add3
> .const .Sub add3 = "add3"
> $P1 = newclosure add3
> $P1()
> .end
>
> .sub add3 :anon :outer(do_add3)
> .return ($P1)
> .end
>
> compile error? runtime error? warning only, and error on lexical
> access in inner sub?
I'd call this correct but silly, kind of like assigning to a register
that doesn't get used afterwards.
--
Chip Salzenberg <[EMAIL PROTECTED]>
