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?
Solution
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).