zing, the short answer is to print the user prompt to STDERR, and redirect STDOUT as you suggest above.

However, there's a more convenient way, if you're up to it. In your OP, you have the command line:

perl short_path.pl input_fie output_file
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but the input and output filenames are never used in the script. To make use of them, you would access @ARGV, either explicitly or implicitly, instead of asking for the input filename.

To do so, you delete this:

print "Enter file name:\n";
chomp(my $file = <STDIN>);
open(DATA,$file) or die "failed opening file!!";
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Where you have:

ins split /,\s*/ for <DATA>;
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replace it with this (based on Athenasius's edit):

for (<>) {
    chomp;
    ins split /,\s*/;
}
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The <> is Perl DWIMmery for the magic <ARGV> filehandle, which steps through all the values in @ARGV as filenames to read from, and does the open for you. (If you expect to feed the script a list of files, you may want to add some user-friendly error checking.)

Then to execute the script:

perl short_path.pl input_file > output_file
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