I use linux. In one script ("/home/weigand/tmp.pl"), I do this:

  #!/bin/env /home/weigand/perl_wrapper.pl
  print "Hello world from perl.\n";

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My "/home/weigand/perl_wrapper.pl" script is just a perl script:

  #!/usr/bin/perl
  print "In perl wrapper.\n";
  ...

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Question: While in the perl_wrapper.pl script, is there any way to get the name of the file "/home/weigand/tmp.pl" in this case?

The reason why I want to know is that I want the perl_wrapper.pl script to execute "perl /home/weigand/tmp.pl". It's a wrapper script for perl, so its job is to determine which perl binary to use and then execute the original script (/home/weigand/tmp.pl in this case) with the right perl binary.

Yes I know there are other ways of writing wrapper scripts. I'd like to stick with this particular question rather than solicit alternatives.

Yes, I know you can make symbolic links at /usr/local/bin to point to different perl binaries for different architectures. That's not what I want.

Any ideas?

I guess I could find the process ID of the parent (which is the /bin/env process I suppose). And then look at its open file list or something. Not sure exactly how to do that or if that's really my best shot.

In reply to Getting the name of the file passed as STDIN from /bin/env? by weigand

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