Function doesn’t update global array

Question

I wrote a simple quick sort algorithm. When I run it, it echoes the same array back.

#!/usr/bin/bash

arr=(1 8 3 9 4 5 7 2)

partition() {
    local low=$1
    local high=$2
    local pivot=${arr[low]}
    local i=$((low - 1))
    local j=$((high + 1))
    local temp

    while true; do
        ((i++))
        while ((arr[i] < pivot && i <= high)); do
            ((i++))
        done

        ((j--))
        while ((arr[j] > pivot && j >= 0)); do
            ((j--))
        done

        if ((i >= j)); then
            echo "$j"
            return
        fi

        temp=${arr[i]}
        arr[i]=${arr[j]}
        arr[j]=$temp
    done
}

quick_sort() {
    local low=$1
    local high=$2

    if ((low >= high)); then
        return
    fi
    
    local pivot
    pivot=$(partition "$low" "$high")

    quick_sort "$low" "$pivot"
    quick_sort $((pivot + 1)) "$high"
}

quick_sort 0 $((${#arr[@]} - 1))

echo "Sorted array: ${arr[*]}"

Output:

Sorted array: 1 8 3 9 4 5 7 2

I am on Windows so i tried it on git-bash and an online bash shell, the outcome was still the same

Your array is not being updated because pivot=$(partition …) runs partition in a subshell. Any changes to arr inside that subshell are lost when it returns, which is why the array looks unchanged. To fix this issue, call partition directly so it mutates the global array, and only return the pivot index (writing it into REPLY or a global variable). Just for exxample: ` partition() { … REPLY=$j; return; } partition "$low" "$high" pivot=$REPLY ` That way the swaps in partition persist and your quick sort works. — Kamran
– Kamran, Commented Sep 17 at 9:47

choroba · Accepted Answer · 2025-09-17 09:49:19Z

3

This line runs partition in a subshell.

pivot=$(partition "$low" "$high")

Variables aren't propagated from a subshell back to the parent shell.

You can insert

set -xv

at the beginning of the script to see what exactly the shell sees and executes.

answered Sep 17 at 9:49

choroba

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1 Comment

Barmar Sep 17 at 17:48

This answer would be improved if it explained how to solve the problem, rather than just giving the reason why it fails.

Ed Morton · Accepted Answer · 2025-09-17 16:39:34Z

Your function is running in the subshell created by $(...) and so cannot change the value of a variable in the shell it's called from. Here's an example of the problem:

$ cat tst.sh
#!/usr/bin/env bash

var=9

foo() {
    var=12
    echo '27'
}

ret=$(foo)

declare -p ret >&2
declare -p var >&2

$ ./tst.sh
declare -- ret="27"
declare -- var="9"

Note that var didn't retain its modified value (12) in the calling shell after foo() returned.

If you can modify foo() then you can get around this problem by creating a nameref variable that refers to ret so you don't need to call foo() in a subshell to populate ret with it's output:

$ cat tst0.sh
#!/usr/bin/env bash

var=9

foo() {
    local -n loc_ret="$1"
    var=12
    loc_ret=27
}

foo ret

declare -p ret >&2
declare -p var >&2

$ ./tst0.sh
declare -- ret="27"
declare -- var="12"

If you cannot modify foo() then here's a solution using a coprocess which doesn't require you to modify the function foo():

$ cat tst1.sh
#!/usr/bin/env bash

var=9

foo() {
    var=12
    echo '27'
}

coproc CAT { cat; }
foo >&${CAT[1]}
IFS= read -r ret <&${CAT[0]}

declare -p ret >&2
declare -p var >&2

$ ./tst1.sh
declare -- ret="27"
declare -- var="12"

In the last script above foo() is being run in the current shell instead of in a subshell and it's output is written to the coprocess we created named CAT, then we read its output back into the variable ret. With this approach var does retain its updated value in the calling shell after foo() returns.

You could, of course, use a temp file instead of a coprocess if you prefer:

$ cat tst2.sh
#!/usr/bin/env bash

var=9

foo() {
    var=12
    echo '27'
}

tmp=$(mktemp) || exit 1
trap 'rm -f "$tmp"; exit' EXIT
foo >"$tmp"
IFS= read -r ret <"$tmp"

declare -p ret >&2
declare -p var >&2

$ ./tst2.sh
declare -- ret="27"
declare -- var="12"

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