Dynamically assign array to value of a variable. Eval?

Question

I'm using Bash 4.

I'm trying out a few experiments with Bash. I want to dynamically assign an array to the value of a variable. If you read the code below, it will be easier to understand.

myFunc =RESULT_PREF 'bar' 'baz'

function myFunc() {
    local params
    parser params "$@"
}


function parser() {
    if [ '=' = "${2::1}" ]; then
        local name="${1}"
        shift 2
        local param_array=( "$@" )

        # In PHP, I'd do:
        # $$name = $param_array

        # The ideal outcome:
        # $params should contain the values 'bar' and 'baz'.
    fi
}

How do I do what I've mentioned in the comments of my code? I can't seem to do this using eval. Even if I could, I've read so many bad things about eval.

Any advice?

Answer 1

In bash 4.3+, there is a way to define a variable reference which you can see in help declare:

-n: make NAME a reference to the variable named by its value

Therefore a script like this outputs a line a:

n(){ declare -n g="$1"; g=a; }
n b; printf '%s\n' "$b";

So here's how you should do that in parser:

parser() {
    [ "$1" ] || ! echo 'no name given' >&2 || return 42
    if [[ "$2" == =* ]]; then
        declare -n name="$1"
        shift 2
        name=("$@")
    fi
}

Using eval does work, but it can be unsafe if you expand the rhs too early:

shopt -e extglob # bash >= 3
[[ "$1" == [[:alpha:]_]*([[:word:]]) ]] || return 42 # invalid varname
# mikeserv has another solution for filtering, basically expanding
# the inverse of the expression to basic, portable globs
# Do this:
eval "$1"'=( "$@" )'
eval "$foo=\$bar"
# But not:
eval "$1=(\"$@\")"
eval "$foo=$bar"

I think you will know what to do with eval then. But simply for a higher possibility that you will accept my answer, let me write it in full:

shopt -s extglob
parser() {
    [[ "$1" == [[:alpha:]_]*([[:word:]]) ]] || ! echo 'bad varname' >&2 || return 42
    if [[ "$2" == =* ]]; then
        name="$1"
        shift 2
        eval "$name=(\"\$@\")"
    fi
}

Answer 2

myFunc(){
    parse "$1" &&                          #parse() is a test
    eval "  shift                          #$1 is a valid, eval-safe name
            local name$1 ${1#=}="'("$@")'  #$1 is expanded before shift
}

parse()
    case ${1#=} in
    ("$1"|""|[0-9]*|*[!_[:alnum:]]*) ! :   #return false for all invalid names
    esac

It does also work without eval given another execution step or two:

myFunc(){
    local -a "name$1[@]" "${1#=}"='("${@:2}")' &&
    [ -z "${1%=*}" ] && printf %s\\n "${!name}"
}

In that case I just allow the local builtin to do all of the validation and assignment at once. Basically, if the local command can assign "${1#=}"=("${@#"$1"}") successfully and "${1%=*}" is null then you know the assignment has taken place successfully. Any syntax error - such as bad shell names - will automatically fail and return with error during the local assignment, and the simple validation [ test ] which follows is all that is necessary to ensure that you don't accidentally do local name=morename=("${@#"$1"}").

The natural upshot to this is that when a bad name is passed in as =$1 the shell will automatically print out a meaningful error message to stderr for you and do all of the error handling automatically with no fuss.

Like this:

myFunc =goodname 1 2 3 'and some     ;more' &&
myFunc =bad-name 1 2 3 'and some     ;more'

---

1
2
3
and some     ;more
bash: local: `bad-name=("${@:2}")': not a valid identifier

---

Note that doing functionname(){list;} is almost definitely not what you mean to do.

If you intend to localize the function's traps and similar you need functionname{list;}.

If you intend to share all state but your locally defined variables with the current shell then functionname(){list;} or name()list are equivalent because the function keyword is ignored by the shell (except that some shells which implement it tend to parse the following list incorrectly if it is not contained in curly braces) when name is followed by ().