You need to make sure you understand defined, exists, and false values. You can enter a value of undef for a given key in a hash. 
my %hash = ();
$hash{'bar'} = undef;

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At this point, defined $hash{'bar'} is false, but exists $hash{'bar'} evaluates to true. defined will signal if a key has a "defined" value. exists signals whether the key "exists". 
$hash{'bar'} = '';

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The above code will evaluate to true when tested with defined or exists. One thing to remember is that there are four false values: '',0,'0', and undef. If you assign undef to a hash key the following if statements are equivalent: 
$hash{'bar'} = undef;

if(defined $hash{'bar'}){print "false"}
if($hash{'bar'}){print "false"}

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If you assign any other false value to a hash key you get a different result: 
$hash{'bar'} = '0';
if(defined $hash{'bar'}){print "true")
if($hash{'bar'}){print "false")

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The lesson, I guess, is when you are testing whether a value has been set to undef you should use defined;it will return false for the value undef, and true for all other values. When testing whether a key exists in a hash then you use exist. When testing whether a value is false or true you should just use the value itself.

In reply to Re: Uninitialized errors when using 2 hashes. by shevek
in thread Uninitialized errors when using 2 hashes. by Anonymous Monk

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