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.

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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).

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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

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

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

Output:

text
Copy
|test test|

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

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

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

Output:

text
Copy
|test test|

Option 3: Avoid the pipeline entirely

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

Output:

text
Copy
|test test|

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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.