How do I get the name of the file being read in Perl?

Question

In the following Perl pattern:

while(<>) {
  # do stuff
}

is there a way to get the name of the file that is presently open?

Just to be clear, I expect to receive many args, so that loop will process more than one file. I want the name of the file presently being processed.

Answer 1

It is stored in

$ARGV

See perldoc perlvar:

• $ARGV

It contains the name of the current file when reading from <>.

However, if are piping in from STDIN, you will get only '-'.

There is also more discussion on the null filehandle in perldoc perlop.