It doesn't start with it's own version of $n, because at the time you're creating $ref1 and $ref2 $n is the same variable.

Doh! Thanks Joost (and imp). :) 

#!/usr/bin/perl

use strict;
use warnings;

my $ref1;
my $ref2;

{
    my $n = 10;
    $ref1 = sub { print( $n++, "\n" ) };
}

{
    my $n = 10;
    $ref2 = sub { print( $n++, "\n" ) };
}

$ref1->(); # --> 10
$ref1->(); # --> 11
$ref1->(); # --> 12

print "\n";

$ref2->(); # --> 10
$ref2->(); # --> 11
$ref2->(); # --> 12

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In reply to Re^4: Using an "outer" lexical in a sub? by cornballer
in thread Using an "outer" lexical in a sub? by cornballer

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