Get filename of STDOUT

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

If I call a script from the command line as follows:

perl process.pl input.txt > output.txt
[download]

...the name of the perl script being executed is in $0 and the name of the input file can be easily extracted from @ARGV. But is there any way of getting at the output filename? E.g.

while (<>) {
    print "Reading from: $ARGV, Writing to: $???\n";
}
[download]

I understand that under normal circumstances it is not possible to extract a filename from a filehandle - but if the name is being passed via the command line as here, is there (or should there be?) a convenient way of retrieving it? _{janitored by ybiC: Retitle from "Missing special variable?" for searchability}

Comment on Get filename of STDOUT Select or Download Code

Replies are listed 'Best First'.
Re: Get filename of STDOUT by Joost (Canon) on Aug 10, 2004 at 17:37 UTC
On unix the command line is parsed by the shell first, so your program only sees the "input.txt" part, and your perl code only has the file handle. A recent discussion in the ChatterBox on the subject brought up the `lsof` Unix command, but I wouldn't call it convenient, and usually needs root permissions. Another possibilty is to use `stat` to get the Inode number and then search the filesystem for the corresponding file. I don't know if that works for multiple file-systems, though. Can't you rewrite your program as taking 2 filenames as arguments? It seems the most convenient by far. J. "What should it profit a man, if he should win a flame war, yet lose his cool?"	[reply] [d/l] [select]
Re: Missing special variable? by etcshadow (Priest) on Aug 10, 2004 at 17:52 UTC
Well, it's highly non-portable, but on linux, you can find this by looking on the proc filesystem and traversing a symlink. You can find the file name of the file to which STDOUT is attached by `readlink "/proc/$$/fd/1";`. Demonstrated as: `[me@host]$ perl -le 'my $outputfile = readlink "/proc/$$/fd/1"; print +STDERR $outputfile;' > t1 /home/me/t1 [me@host]$` [download] That is... look into this process (/proc/$$), examine its filehandles (/proc/$$/fd). In particular, look at its filehandle number 1 (the fileno of STDOUT). This will be a symlink to the actual file that STDOUT is opened to (hence the readlink). `------------ :Wq Not an editor command: Wq` [download]	[reply] [d/l] [select]
Re: Get filename of STDOUT by pelagic (Priest) on Aug 10, 2004 at 17:35 UTC
I have never tried out but you might find some solution in here: filenames from filehandles pelagic	[reply]
Re: Get filename of STDOUT by gaal (Parson) on Aug 10, 2004 at 17:42 UTC
Bad news: even `$0` won't work on all platforms. Win32, notably, does not support it.	[reply] [d/l]
Re^2: Get filename of STDOUT by eserte (Deacon) on Aug 10, 2004 at 18:05 UTC
Bad news: even $0 won't work on all platforms. Win32, notably, does not support it. ActivePerl 808 on Windows seems to support $0.	[reply]
Re^3: Get filename of STDOUT by gaal (Parson) on Aug 10, 2004 at 18:23 UTC
Really? What does it say? (For named scripts; for one-liners?) Can you modify it? Last time I checked it said something useless like "perl.exe" or maybe nothing at all. That was in some 5.6 ActivePerl.	[reply]
Re^4: Get filename of STDOUT by Ven'Tatsu (Deacon) on Aug 10, 2004 at 19:20 UTC
Re^5: Get filename of STDOUT by gaal (Parson) on Aug 11, 2004 at 07:41 UTC
Re^4: Get filename of STDOUT by ikegami (Patriarch) on Aug 10, 2004 at 20:21 UTC