How can I pass an argument in a shell script without quotes?

kernel_kate

I have to make my own command shell where a bash script calls the shell executable and does so with quotes. The problem is that when I try to do so with logical operators it seems that the quotes turn off the special characters && . For example: var=("echo hello && mkdir testing") ./rshell ${var} The command shell ends up outputting hello && mkdir testing instead of actually following the logic of the operator where it successfully outputs hello...

Originally asked by Code4life on Stack Exchange (unix) (CC BY-SA 4.0).

uphilltech

Great question — answering as the UphillTech team:

This definitely seems like an x-y problem . It would help to know what the ultimate end goal for this is, but: Your variable var is actually an array because of the var=( ... ) syntax Because you quote everything in the parenthesis you are adding all data to a single element You are not using the proper syntax to call an array ( ${var[@]} ) which in this example isn't a big deal since everything is in a single element anyway The biggest problem seems to be the way the logical AND gets quoted and interpreted. If you don't quote it when adding to the array the shell will attempt to evaluate it, and if you do quote it, it will remain a literal string preventing evaluation. One of the ways I...

Full answer by jesse_b: https://unix.stackexchange.com/a/518413 (CC BY-SA 4.0).

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