Writing an script interpreter in bash

Question

I wrote a bash wrapper to ruby that goes through various setup steps.

The most basic version is,

   #!/bin/bash 
   #   ruby_wrapper.sh
   ruby

Now I want to be able to use this wrapper just like regular ruby! Specifically, I want to create a .rb file that uses this "interpreter".

   #!/path/to/ruby_wrapper.sh
   #  my_file.rb
   puts "hello world"

So I want to be able to do $ ./my_file.rb instead of $ ruby_wrapper.sh my_file.rb

Is this possible?

The documentation claims it isn't.

Note that the interpreter may not itself be an interpreter script.

But I don't see why not. Does anyone have any ideas to get around this?

Answer 1

Try invoking your wrapper with /usr/bin/env . It's actually good practice to execute Ruby scripts with /usr/bin/env ruby as you don't have to hard code the path to the ruby binary, so this is not unnatural.

$ cat ruby_wrapper.sh 
#!/bin/bash
exec ruby "$@"

$ cat wrapped.rb 
#!/usr/bin/env /tmp/ruby_wrapper.sh
puts "hello world"

$ ./wrapped.rb 
hello world

Also as a side note see how I've used exec in the wrapper script. This will allow the ruby interpreter to take over your wrapper script's process rather than run as a child process of your script. This makes the wrapper transparent to the caller (signals will be delivered directly to ruby rather than to bash, for example).