Sadly, it's a feature, but it's the intermediate keys you are creating. Try this: 
    print "1st run\n" if exists $a{a};
    print "2nd run\n" if exists $a{a}{b}{c};
    print "3rd run\n" if exists $a{a};

# prints 3rd run

[download]
$a{a}{b} is created in order to test the existence of $a{a}{b}{c}. From perldoc -f exists: 
Note that the EXPR can be arbitrarily complicated as long as the
final operation is a hash or array key lookup or subroutine
name:

    if (exists $ref->{A}->{B}->{$key})  { }
    if (exists $hash{A}{B}{$key})       { }

    if (exists $ref->{A}->{B}->[$ix])   { }
    if (exists $hash{A}{B}[$ix])        { }

    if (exists &{$ref->{A}{B}{$key}})   { }

Although the mostly deeply nested array or hash will not spring
into existence just because its existence was tested, any
intervening ones will. Thus "$ref->{"A"}" and
"$ref->{"A"}->{"B"}" will spring into existence due to the
existence test for the $key element above. This happens anywhere
the arrow operator is used, including even here:

    undef $ref;
    if (exists $ref->{"Some key"})    { }
    print $ref;  # prints HASH(0x80d3d5c)

This surprising autovivification in what does not at first--or
even second--glance appear to be an lvalue context may be fixed
in a future release.

[download]
But you can be smart about this. You can test for $a{a}. If it doesn't exist, any and all subkeys certainly won't exist.

--marmot

In reply to Re: Why does perl expression in empty nested hash create parents by furry_marmot
in thread Why does perl expression in empty nested hash create parents by Steve_BZ

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