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.

Accepted Answer

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

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

Solution

Problem