Using arrays in a shell script

Question

I am trying to initialise an array in a shell script like below

declare -a testArr=( 'A' 'B' )

Also like below

testArr=( 'A' 'B' )

But in both the cases, I get the following error

shell1.sh: 1: shell1.sh: Syntax error: "(" unexpected

Can someone please tell me the reason for the above error?

Very likely your shebang is #!/bin/sh and not #!/bin/bash. Arrays are not supported in POSIX shells, but they are in Bash. — gniourf_gniourf
– gniourf_gniourf, Commented Dec 16, 2014 at 10:39
@gniourf_gniourf Ya sorry it is not bash. How can i use arrays in POSIX shells? — Aarish Ramesh
– Aarish Ramesh, Commented Dec 16, 2014 at 10:41
you can't. sh doesn't support arrays. If you need arrays, switch to Bash. — gniourf_gniourf
– gniourf_gniourf, Commented Dec 16, 2014 at 10:42

drizzt · Accepted Answer · 2014-12-16 11:04:49Z

3

Since you want to use POSIX shell and POSIX shell does not support arrays, you can "emulate" array in this way:

set -- 'A' 'B'

Then you will have the only "array" available in POSIX shell ("$@") that contains A ($1) and B ($2).

And you can pass this array to another function, such as:

test() {
    echo "$2"
}

set -- 'A' 'B C'
test "$@"

If you need to save the array you can use the following function:

arrsave() {
    local i
    for i; do
        printf %s\\n "$i" | sed "s/'/'\\\\''/g;1s/^/'/;\$s/\$/' \\\\/"
    done
    echo " "
}

And use it in the following way:

set -- 'A' 'B C'

# Save the array
arr=$(arrsave "$@")

# Restore the array
eval "set -- $arr"

answered Dec 16, 2014 at 11:04

drizzt

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