CODEWITHSUNDEEP
bash
Cinolt Yuklair
Cinolt Yuklair

Reputation: 361

Passing arguments to another script, after having used the "shift" builtin

Consider a simple script program.sh:

$ cat program.sh
echo program: I have "$#" arguments, they are "$@"
$ ./program.sh 1 2 3
program: I have 3 arguments, they are 1 2 3
$ ./program.sh '1 1' 2 3
program: I have 3 arguments. they are 1 1 2 3

Now I want a wrapper script wrapper.sh that invokes program.sh in the same way it was invoked:

$ cat wrapper.sh
echo wrapper: I have "$#" arguments, they are "$@"
./program.sh "$@"
$ ./wrapper.sh 1 2 3
wrapper: I have 3 arguments, they are 1 2 3
program: I have 3 arguments, they are 1 2 3
$ ./wrapper.sh '1 1' 2 3
wrapper: I have 3 arguments. they are 1 1 2 3
program: I have 3 arguments. they are 1 1 2 3

This works perfectly, but I want to use the shift builtin within wrapper.sh, which creates a problem because then I cannot use "$@" anymore in the invocation to program.sh:

$ cat wrapper.sh
echo wrapper: I have "$#" arguments, they are "$@"
shift
./program.sh "$@"
$ ./wrapper.sh 1 2 3
wrapper: I have 3 arguments, they are 1 2 3
program: I have 2 arguments, they are 2 3
$ ./wrapper.sh '1 1' 2 3
wrapper: I have 3 arguments. they are 1 1 2 3
program: I have 2 arguments. they are 2 3

I have tried initially setting the value of $@ to a temporary variable, but nothing I tried seems to work for wrapper.sh to invoke program.sh exactly the way it was originally invoked. What is the proper way to handle this?

Upvotes: 2

Views: 36

Answers (1)

John1024
John1024

Reputation: 113864

It is easy to save $@. Just save it as an array:

$ cat wrapper.sh 
echo wrapper: I have "$#" arguments, they are "$@"
save=("$@")
shift
./program.sh "${save[@]}"

This produces the output:

$ wrapper.sh '1 1' 2 3
wrapper: I have 3 arguments, they are 1 1 2 3
program: I have 3 arguments, they are 1 1 2 3

The key here is that save is a bash array. Trying to store $@ as a shell string does not work. Saving it as an array does work.

Upvotes: 4

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