XK0-006 Automation, Orchestration, and Scripting Practice Question
In a Bash script, a variable is assigned the output of a command using: result=$(ls -l). What is the purpose of the $() syntax?
⚠ Common exam trap
The trap is confusing command substitution $() with variable expansion ${}, causing candidates to pick the 'expands the variable' distractor.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
It runs the command and assigns its output to the variable
The $() syntax is command substitution: Bash executes the command inside the parentheses in a subshell, captures its standard output, and substitutes that text into the assignment. So 'result=$(ls -l)' stores the directory listing in the variable result. This is the modern, nestable form of the older backtick syntax.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
It runs the command and assigns its output to the variable
Why this is correct
Command substitution executes the enclosed command in a subshell and replaces the expression with its standard output, which is then assigned to the variable. This lets result capture the directory listing text rather than the literal string, satisfying the script's intent.
- ✗
It expands the variable result
Why it's wrong here
Command substitution captures the command's standard output and substitutes it into the assignment; it does not expand the variable result, which is the assignment target. Variable expansion is performed by the $result or ${result} syntax, not by $().
- ✗
It checks if the command exists
Why it's wrong here
Command substitution captures stdout into the variable; it never tests command existence. Testing existence uses command -v or type. The syntax is tempting because $(...) superficially resembles a conditional construct, but it performs substitution, not validation.
- ✗
It runs the command in a subshell and discards output
Why it's wrong here
Command substitution runs the command in a subshell but captures its stdout into the variable rather than discarding it. Discarding output requires redirection to /dev/null. The subshell execution is tempting because $(...) genuinely forks a subshell, yet capture, not disposal, is the purpose.
Visual reference
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Related to this question
Key term
Output
In IT service management, output is the result or deliverable produced by a process, system, or component, such as data, reports, or services delivered to a customer.
Key term
Bash script
A Bash script is a text file containing a sequence of commands for the Unix shell Bash, allowing users to automate repetitive tasks and streamline system administration on Linux and macOS.
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JA
Written and reviewed by Johnson Ajibi, MSc IT Security
Senior Network & Security Engineer · founder of Courseiva
Last reviewed September 2026 · checked against the official CompTIA exam blueprint
This XK0-006 practice question is part of Courseiva's free CompTIA certification practice question bank. Courseiva provides original exam-style practice questions with explanations, topic-based practice, mock exams, readiness tracking, and study analytics to help learners prepare for the XK0-006 exam.