The docs say

If the filename contains no directory components XMLin() will look for the file in each directory in the SearchPath (see "OPTIONS" below) or in the current directory if the SearchPath option is not defined.

The search path is initialized to something based on $0. In this case, it's incorrectly initialized to ['']. Try 

$data = $xml->XMLin($file, SearchPath => '.');

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In reply to Re: XMLin is not identifying the input file name passed to it by ikegami
in thread XMLin is not identifying the input file name passed to it by greatshots

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