bash - pass script as argument of another script

Question

I can't find a similar question on SO.

How can I properly pass a bash script as an argument to another bash script.

For example, let's say I have two scripts that can each accept a number of parameters, I want to pass one script as the argument of the other. Something like:

./script1 (./script2 file1 file2) file3

In the above example, script2 merges file1 and file2 together, and echos a new file, however that is irrelevant to the question. I just want to know how I can pass script2 as a parameter, i.e. the proper syntax.

If this is not possible, any hint as to how I may circumvent the issue would be appropriate.

Answer 1

If you want to pass the result of evaluating script2 as a parameter, use $(). Remember that you have to quote it.

./script1 "$(./script2 file1 file2)" file3