Check if Array1 length is equal to Array2 length (bash)

Question

In a for loop I am looking to find out: If Array1 length matches Array2 length then break the for loop.

Shellcheck throws an error (while the script runs fine)

    if [[ "${!Array1[@]}" == "${!Array2[@]}" ]] ; then
            break;
    fi

^-- SC2199: Arrays implicitly concatenate in [[ ]]. Use a loop (or explicit * instead of @).

I'm still learning bash and my teacher said "Always verify with Shellcheck" and to "always place conditionals in double [[ ]]" and to "never use * for array length"

The error(s) are removed with the correction

    if [ "${!Array1[*]}" == "${!Array2[*]}" ] ; then
            break;
    fi

I was wondering what is the best practice here?

If by length you mean the number of elements, "${!Array1[@]}" is wrong. — Cyrus
– Cyrus, Commented Sep 25, 2017 at 7:35
Obtain the length of (count of elements in) an array with "${#array[@]}". — Leon
– Leon, Commented Sep 25, 2017 at 7:37

nyronium · Accepted Answer · 2017-09-25 08:00:27Z

1

Your code is partly correct. The problem is, that you are expanding the array's indicies with th ! operator, and not the length by using the # operator.

Thus a warning about implicit concatenation is issued for both uses of ${array[@]}, as the list of indicies is 0 1 2 3 .... Nevertheless your code is working, as two non associative bash arrays with equal length will have identical index lists 0 1 2 .. N.

To get rid of the warning, you should replace ${!array[@]} with ${#array[@]}. Of course, using ${!array[*]} will also suppress the warning, but that is definitely not what you want to do here, as you would continue comparing index lists.

For further reading: The author of shellcheck has explained the concatenation issue here in detail.

edited Sep 25, 2017 at 8:00

answered Sep 25, 2017 at 7:42

nyronium

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1 Comment

Leon Over a year ago

two non associative bash arrays with equal length will have identical index lists 0 1 2 .. N This is generally not true, as there may be gaps in the used values of indices, e.g. declare -a a='([0]="a" [1]="b" [10]="c")' for which ${#a[@]} evaluates to 3 and ${#a[@]} is 0 1 10.

Ben · Accepted Answer · 2017-09-25 11:30:22Z

Nyronium's works well and explains the solution perfectly.

I also found a solution that utilities # to check the array length for the condition.

Sorry I did not have this example script before, as I wrote it to test logic afterwards.

Example

#!/bin/bash
array1=(1 2 3)
array2=()

echo "array1 length: ${#array1[@]}" #output :3

for i in {1..10} ; do
        echo "array2 length: ${#array2[@]}"
        array2+=("$i")

        #I want to exit when array2 = a length of 3
        if [[ "${#array1[@]}" == "${#array2[@]}" ]] ; then
                echo "exit"
                break
        fi
done

echo "final array1 length: ${#array1[@]} vs array2 length: ${#array1[@]}"

Result: final array1 length: 3 vs array2 length: 3

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