Subroutine giving two different outputs when called twice [duplicate] - perl

Question

This question already has an answer here:
Perl - What scopes/closures/environments are producing this behaviour?
(1 answer)
Closed 5 years ago.
I have executed the following piece of a simple nested subroutine and the output of it makes me crazy.
#!/usr/bin/perl
use strict;
use warnings;
sub outer {
my $a = "123";
sub inner {
print "$a\n";
}
inner();
$a = "456";
}
outer();
outer();
Output to this is
Variable "$a" will not stay shared at E:\Perl\source\public\sss.pl line 9.
123
456
But how is this possible?
I call the inner subroutine when $a value is 123, but why am I getting 456 when outer is called the second time.

Answer 1

perldoc diagnostics gives quite self explanatory description for the warning Variable "$a" will not stay shared,
use strict;
use warnings;
use diagnostics;
sub outer {
my $a = "123";
sub inner {
print "$a\n";
}
inner();
$a = "456";
}
outer();
outer();
output
Variable "$a" will not stay shared at -e line 9 (#1)
(W closure) An inner (nested) named subroutine is referencing a
lexical variable defined in an outer named subroutine.
When the inner subroutine is called, it will see the value of
the outer subroutine's variable as it was before and during the *first*
call to the outer subroutine; in this case, after the first call to the
outer subroutine is complete, the inner and outer subroutines will no
longer share a common value for the variable. In other words, the
variable will no longer be shared.
This problem can usually be solved by making the inner subroutine
anonymous, using the sub {} syntax. When inner anonymous subs that
reference variables in outer subroutines are created, they
are automatically rebound to the current values of such variables.
123
456

Answer 2

There is no point in declaring an subroutine within another one. It works as if it were declared at the top level, and won't function properly as a closure
If you enable lexical subroutines (and disable the corresponding "experimental" warning) and declare inner as my sub inner then your code will work as you expect
#!/usr/bin/perl
use strict;
use warnings 'all';
use feature 'lexical_subs';
no warnings 'experimental::lexical_subs';
sub outer {
my $a = "123";
my sub inner {
print "$a\n";
}
inner();
$a = "456";
}
outer();
outer();
output
123
123