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The stupid question is the question not asked

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if (($english && $foreign) eq '-1') {

The expression $english && $foreign will evaluate to either true (1) or false ("") but will never be equal to '-1'.

sub hash_processor::function () { my ($english, $foreign, $process) = @_;

Your prototype says that the subroutine accepts no arguments but the next line says that your subroutine does accept three arguments.    You shouldn't use prototypes.

You are using package names but I don't see where you are creating a package namespace.

if ($process == '0') {

That is usually written as $process == 0 or $process eq '0' depending on whether $process contains a string or a number.

$english = "$english" . "$hash_processor::tag_count"; $foreign = "$foreign" . "$hash_processor::tag_count";

That is usually written as:

$english .= $hash_processor::tag_count; $foreign .= $hash_processor::tag_count;

In reply to Re: Hash Printing reveals Hash Memory Location by jwkrahn
in thread Hash Printing reveals Hash Memory Location by perlStuck

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