bash argument parsing in function

Question

Is the a workaround to parsing the bash script arguments in an function

run command: ./script.sh -t -r -e

script:

#!/bin/sh
# parse argument function
parse_args() {
echo "$#"   #<-- output: 0
}


# main
echo "$#"   #<-- output: 3

# parsing arguments
parse_args

I now the reason for the 'output 0' but is there a way to do the argument parsing in an extra function? — Heinrich
– Heinrich, Commented Sep 6, 2014 at 17:37

David Foerster · Accepted Answer · 2020-02-27 12:34:50Z

11

$# evaluates to the number of parameters in the current scope. Since each function has its own scope and you don't pass any parameters to parse_args, $# will always be 0 inside of it.

To get the desired result, change the last line to:

parse_args "$@"

The special variable "$@" expands to the positional parameters of the current (top-level) scope as separate words. Subsequently they are passed to the invocation of parse_args.

edited Feb 27, 2020 at 12:34

answered Sep 6, 2014 at 17:50

David Foerster

1,5511 gold badge14 silver badges25 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

GhostCat Over a year ago

Nice answers. And great example you added to my answer!

Collectives™ on Stack Overflow

bash argument parsing in function

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related