Sorry, I used the wrong var name. What I meant is that fixing the failing case below and will cause the third to fail.

uuse strict;
use warnings;
no strict 'vars';

use Test::More tests => 3;

sub var_exists {
   my ($var) = @_;
   return defined ${substr $var, 1};
}

{
   no warnings 'once';
   $var1 = 'a';
   $var2 = undef;
   @var3 = qw( a b c );
}

ok(var_exists('$var1'),  'var1');
ok(var_exists('$var2'),  'var2');
ok(!var_exists('$var3'), 'var3');
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1..3
ok 1 - var1
not ok 2 - var2
#   Failed test 'var2'
#   at a.pl line 20.
ok 3 - var3
# Looks like you failed 1 test of 3.
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From a Perl (language!) point of view, there's no difference between a package variable that hasn't been used, and a package variable which is undefined. The following, after all, is a valid Perl program; it compiles and runs:

   print $foo;
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it may warn, but it's valid.

It only matters in the implementation, and in the few edge cases where you can poke into the underlaying implementation.

So why bother using use strict; if it doesn't matter whether a variable exists or not. That perl doesn't care says nothing about whether the programmer cares or not.

Note I never said that the OP cares. I even said he might not ("I don't know if that's ok or not."). I'm just outlining the possibilities the OP might be interested in.