instead of
copy( "$dir\\$file", "C:\\testfiles\\$file" ) or die "Failed to copy $file: $!\n"; rename $file, $newfile
You should try:
copy( "$dir\\$file", "C:\\testfiles\\$file" ) or die "Failed to copy $file: $!\n"; rename "C:\\testfiles\\$file", "C:\\testfiles\\$newfile";
But have you ever tried this:
copy( "$dir\\$file", "C:\\testfiles\\$newfile" ) or die "Failed to copy $file: $!\n";
Question: What will happen if you copy more than one file from $dir?

Answer: You will end up with just one $newfile.

Then regarding your not-shifted $newfile parameter to your sub. Instead of shifting a sub's parameters I prefer to write:

sub process_dir { my ($dir, $newfile) = @_;

In reply to Re: Copy and Rename file at destination directory by Skeeve
in thread Copy and Rename file at destination directory by Skyler99

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