How do I get `__FILE__` when sourcing a csh script

I have a script that is used to set some env vars in the calling csh shell. Some of those variables depend on the location of the script.

If the file is a proper csh script, I can use $0 to access __FILE__ but if I run the script using source, it just tells me csh or tcsh.

Since I'm using this to set var开发者_开发问答s in the parent shell, I have to use source.

What to do?

If you access $_ on the first line of the file, it will contain the name of the file if it's sourced. If it's run directly then $0 will contain the name.

#!/bin/tcsh
set called=($_)
if ($called[2] != "") echo "Sourced: $called[2]"
if ($0 != "tcsh") echo "Called: $0"

This is hard to read, but is actually works:

If your script is named test.csh

/usr/sbin/lsof +p $$ | \grep -oE /.\*test.csh

this answer describes how lsof and a bit of grep magic is the only thing that seems to stand a chance of working for nested sourced files under csh/tcsh.

If your script is named source_me.tcsh:

/usr/sbin/lsof -p $$ | grep -oE '/.*source_me\.tcsh'

The $$ does not work when one calls the source within a sub-shell. BASH_PID only works in bash.

% (sleep 1; source source_me.csh)

I found the following works a bit better:

% set pid=`cut -d' ' -f4 < /proc/self/stat`
% set file=`/usr/sbin/lsof +p $pid|grep -m1 -P '\s\d+r\s+REG\s' | xargs | cut -d' ' -f9`

First line finds the pid of the current process that is sourcing the file. This came from Why is $$ returning the same id as the parent process?. The second line finds the first (and hopefully only) FD opened for read of a regular file. This came from engtech above.