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Hi, I have a script that needs to find out where it was executed from. The following script:
#!/usr/bin/perl -w
print "$0\n";
normally works ok. However, if the script is set-uid (to any user) it prints something like "/dev/fd/3". This is under Solaris 2.7. The FindBin modules uses $0 as its starting point, so it doesn't work either.

I read somewhere that this is a trick used by some kernels to safely execute set-uid scripts: when a set-uid script is invoked, the kernel passes the script to the interpreter using a file descriptor instead of the file name directly, to avoid race conditions. If this is the case, my problem may be unsolvable, but I thought I would ask anyway.

I asked in comp.lang.perl.misc about this, and got two main suggestions, which I list with their drawbacks:

  • Have a non-suid wrapper script that gets the program path, stores in in an environment variable and then executes the suid script. Drawback: anyone can set the environment variable and execute the suid script directly.
  • Have a "helper" non-suid script that prints its path to stdout, and that is stored in the same directory as the suid script. Then the suid script can use it to find out its path. Drawback: the directory where both scripts are stored has to be in the user's path.
So I'm still looking for a real answer to my question. Any ideas will be very appreciated!



In reply to How to determine the program path from a set-uid program by ZZamboni

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