Unless I'm mistaken, this code won't do anything at all. You're getting each $key directly from the hash, so it's always going to exist. Perhaps you meant if (!defined $my_hash{$key})?

Anyway, checking for defined-ness won't matter either since the "undefined" values are actually empty array refs, which are defined.

In reply to Re^4: Printing only the "keys that have a value" in hash by lostjimmy
in thread Printing only the "keys that have a value" in hash by iphone

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