Perl6 Late variable declaration allowed?

[rir]

use v6;     

say rout();

sub rout { return $var; }
   
my $var = "Hey";

This outputs Any().

Is this correct behavior?

Is there or will there be a uninitialized warning mechanism?

Be well,
rir

[moritz]

It's a bug in Rakudo that it doesn't reject the program at compile time, because $var isn't declared (yet). It's one of our oldest open bugs.

Update: It should be noted though that lexical variables are always scoped to a block in Perl 6, and that the detection of of undeclared variables happens only at compile time. Thus this piece of code works differently in Perl 5 and Perl 6 (and this time Rakudo is correct here):

$ perl -Mstrict -wE 'sub f { say eval q[$x] }; my $x = 5; f'
Use of uninitialized value in say at -e line 1.

$ perl6 -e 'sub f { say eval q[$x] }; my $x = 5; f'
5
$

Here the $x is scoped to the whole mainline (which is an implicit block), so an eval sees it at runtime even though the eval comes textually before the declaration of $x.

Second update: I've fixed the bug, so updating to the latest development version should give the desired behavior.

The fix is a bit incomplete, but you're now much more unlikely to come across this problem.

[zentara]

Sooooo, the Perl6 lessons are beginning to start.:-)