syntax error while working with array

Question

I am creating a function were the first argument would be used as a global variable that holds an array of numbers populated by the rest of the argument but i get a syntax error while assigning the values, but if i just assign a scalar to the variable it works. This is the code below

the value of new_set_name which is a would hold all the array elements

#!/usr/bin/env bash

create() {

    local new_set_name=${1}

    shift;

    declare -g ${new_set_name}=( ${@} );    
    echo ${!new_set_name}
}

create a 1 2 3 4

echo ${a[@]}

but if i tried it with a scalar it works

#!/usr/bin/env bash

create() {

    local new_set_name=${1}

    shift;

    declare -g ${new_set_name}=1;

    echo ${!new_set_name}
}

create a 1 2 3 4

echo ${a[@]}

am a bit surprised to see it work for scalars and errors are spit out for arrays. How can i solve this ?

PesaThe · Accepted Answer · 2018-01-07 18:16:53Z

3

In bash 4.3 or later, you can use nameref:

#!/usr/bin/env bash

create() {
    local -n name=$1; shift
    name=("$@")
}

create a 1 2 3 4
printf '%s\n' "${a[@]}"

See Bash FAQ #6 for more information.

edited Jan 7, 2018 at 18:16

answered Jan 7, 2018 at 18:00

PesaThe

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5 Comments

0.sh Over a year ago

just the answer am looking for thanks, but it seems like there is no documentation for nameref , when i do help -m local , the -n switch was not specified

PesaThe Over a year ago

@0.sh There is a link in the answer. Click on nameref.

PesaThe Over a year ago

@0.sh Also, make sure to check out the updated link. And use help declare instead of help local.

0.sh Over a year ago

oh, i thought local , declare, typeset do the same thing ?

PesaThe Over a year ago

@0.sh The thing is, the help for local just says OPTION can be any option accepted by 'declare'. So to see the options, -n for example, you need to use help declare.

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