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

If you're using linux, you can take a look at the file pointed to by /proc/self/fd/0.

Edit: Answer amended for clarity: The above is useful, but only in cases where input for the perl script is read from stdin. This can be determined by reading $ARGV, as described in the replies above.

Answer 2

It is stored in

$ARGV

See perldoc perlvar:

• $ARGV

  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

Answer 3

$ARGV contains the name of the file used in <>.