in reply to passing filename in subroutines
Either of your examples is potentially correct, though I suspect neigher is what you want. my $filename = aa; will put a value into $filename if you've previously defined a suboutine "aa" that returns a value. my $filename = <aa>; will read a line from a file that you've opened using the symbol "aa" to hold the file descriptor.
I suspect you want my $filename = "foo"; In any case, subroutine($filename); is the correct way to invoke a subroutine, passing whatever $filename holds as the single argument.
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