shell - Is there a way to avoid positional arguments in bash?

Question

I have to write a function in bash. The function will take about 7 arguments. I know that I can call a function like this:

To call a function with parameters:

function_name $arg1 $arg2

And I can refer my parameters like this inside the function:

function_name () {
   echo "Parameter #1 is $1"
}

My question is, is there a better way refer to the parameters inside the function? Can I avoid the $1, $2, $3, .... thing and simply use the $arg1, $arg2, ...?

Is there a proper method for this or do I need to re-assign these parameters to some other variables inside the function? E.g.:

function_name () {
   $ARG1=$1
   echo "Parameter #1 is $ARG1"
}

Any example would be much appreciated.

Answer 1

The common way of doing that is assigning the arguments to local variables in the function, i.e.:

copy() {
    local from=${1}
    local to=${2}

    # ...
}

---

Another solution may be getopt-style option parsing.

copy() {
    local arg from to
    while getopts 'f:t:' arg
    do
        case ${arg} in
            f) from=${OPTARG};;
            t) to=${OPTARG};;
            *) return 1 # illegal option
        esac
    done
}

copy -f /tmp/a -t /tmp/b

---

Sadly, bash can't handle long options which would be more readable, i.e.:

copy --from /tmp/a --to /tmp/b

For that, you either need to use the external getopt program (which I think has long option support only on GNU systems) or implement the long option parser by hand, i.e.:

copy() {
    local from to

    while [[ ${1} ]]; do
        case "${1}" in
            --from)
                from=${2}
                shift
                ;;
            --to)
                to=${2}
                shift
                ;;
            *)
                echo "Unknown parameter: ${1}" >&2
                return 1
        esac

        if ! shift; then
            echo 'Missing parameter argument.' >&2
            return 1
        fi
    done
}

copy --from /tmp/a --to /tmp/b

Also see: using getopts in bash shell script to get long and short command line options

---

You can also be lazy, and just pass the 'variables' as arguments to the function, i.e.:

copy() {
    local "${@}"

    # ...
}

copy from=/tmp/a to=/tmp/b

and you'll have ${from} and ${to} in the function as local variables.

Just note that the same issue as below applies — if a particular variable is not passed, it will be inherited from parent environment. You may want to add a 'safety line' like:

copy() {
    local from to    # reset first
    local "${@}"

    # ...
}

to ensure that ${from} and ${to} will be unset when not passed.

---

And if something very bad is of your interest, you could also assign the arguments as global variables when invoking the function, i.e.:

from=/tmp/a to=/tmp/b copy

Then you could just use ${from} and ${to} within the copy() function. Just note that you should then always pass all parameters. Otherwise, a random variable may leak into the function.

from= to=/tmp/b copy   # safe
to=/tmp/b copy         # unsafe: ${from} may be declared elsewhere

---

If you have bash 4.1 (I think), you can also try using associative arrays. It will allow you to pass named arguments but it will be ugly. Something like:

args=( [from]=/tmp/a [to]=/tmp/b )
copy args

And then in copy(), you'd need to grab the array.