That code, while it found the main code error, will still
sometimes show the error message.
$secretword=$words{$name};
if ($secretword eq ""){
$secretword="stupid";
}
If $name is not found in the hash %words, then
$secretword will be undefined and will produce the error mentioned
when used as if defined. If you however test for the
definedness of $secretword you will be closer to the
original aim.
if (defined $secretword) {
$secretword = 'stupid';
}
In perl style I'd probably write:
$secretword = $words{$name} || 'stupid';
But that is a matter of choice.
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