The issue in your script is due to variable scope in Bash. In Bash, variables declared inside a while loop (or any subshell) are local to that subshell and do not persist outside of it. The while loop in your script is part of a pipeline (echo test | while ...), which runs in a subshell. As a result, the args array is modified inside the subshell, but the parent shell (where you try to print args) does not see those changes.

Why it happens

• The while loop runs in a subshell because it is part of a pipeline (echo test | while ...).
• The args array is modified inside the subshell, but the parent shell retains its original value (empty).

How to fix it

To ensure the args array is modified in the parent shell, avoid using a pipeline. Instead, use here-strings or process substitution to feed input to the while loop without creating a subshell.

Option 1: Use a here-string

args=()
while read option
do
    args+=("$option" "$option")
    break
done <<< "test"

echo "|${args[@]}|"

Output:

|test test|

Option 2: Use process substitution (if you need to read from a command)

args=()
while read option
do
    args+=("$option" "$option")
    break
done < <(echo test)

echo "|${args[@]}|"

Output:

|test test|

Option 3: Avoid the pipeline entirely

args=()
option="test"
args+=("$option" "$option")
echo "|${args[@]}|"

Output:

|test test|

Key Takeaway

• Pipelines create subshells in Bash, and variables modified in a subshell do not affect the parent shell.
• Use here-strings (<<<), process substitution (< <(...)), or direct assignment to avoid subshell issues.