Ahh, the follies of not using strict...

These are the three crucial calls in your routine:

  1. exists $test_hash{$key1}
  2. exists $test_hash{$key1}{$key2}
  3. exists $test_hash{$key1} # again
During the first call, you test whether or not $test_hash contains a key called $key1. Of course it doesn't, so exists returns false.

During the second call, you first grab hash reference stored as a value in $test_hash with key $key. This doesn't exist, so Perl dutifully creates it for you on the fly (the behavior when you don't use strict). You then check to see if this anonymous hash contains a key equal to $key2, which it doesn't.

Since you created an anonymous hash reference on the fly in step 2, $test_hash{$key1} now contains that reference. So exists returns true.

Thus, exists() did not fill the value for $test_hash{$key1}; you implicitly did, when you did a lookup on a hash reference that previously did not exist. This would become apparent had you used strict.

The motto of the story is always use strict!

UPDATE: As MeowChow correctly points out, use strict does not catch this. Neither does -w. Oops. You should still always use strict, though :)

-Ton

-----

Be bloody, bold, and resolute; laugh to scorn
The power of man...

In reply to Re: exists() unexpected behavior by ton
in thread exists() unexpected behavior by Richard

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