- A
set -e; ./risky_command; rc=$?; echo $rc
Why wrong: A is wrong because 'set -e' would cause exit before capturing status.
- B
rc=$? ./risky_command; echo $rc
Why wrong: D is wrong because $? is evaluated before the command runs.
- C
./risky_command; rc=$?; echo $rc
C is correct. The exit status is always captured after command.
- D
./risky_command && rc=$?; echo $rc
Why wrong: B is wrong because 'rc' is only set if the command succeeds.
Question 264 of 537
EX200 Create simple shell scripts Practice Question
This EX200 practice question tests your understanding of create simple shell scripts. Read the scenario carefully and evaluate each option against the stated constraints before committing to an answer. After answering, compare your reasoning against the explanation and wrong-answer breakdown below. Once you have made your selection, read the full explanation to reinforce the concept and understand why each distractor is designed to mislead on exam day.
A script needs to execute a command that might fail, but the script should continue. The administrator wants to capture the exit status for logging. Which code snippet correctly implements this?
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
./risky_command; rc=$?; echo $rc
Option C is correct because it runs the risky command, captures its exit status immediately after in the `$?` variable, and then echoes it for logging. The script continues regardless of the command's success or failure, which meets the requirement. The `$?` variable holds the exit status of the last executed foreground command, so assigning it to `rc` right after `./risky_command` ensures the correct value is stored.
Key principle: Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
set -e; ./risky_command; rc=$?; echo $rc
Why it's wrong here
A is wrong because 'set -e' would cause exit before capturing status.
- ✗
rc=$? ./risky_command; echo $rc
Why it's wrong here
D is wrong because $? is evaluated before the command runs.
- ✓
./risky_command; rc=$?; echo $rc
Why this is correct
C is correct. The exit status is always captured after command.
Related concept
Read the scenario before looking for a memorised answer.
- ✗
./risky_command && rc=$?; echo $rc
Why it's wrong here
B is wrong because 'rc' is only set if the command succeeds.
Common exam traps
Common exam trap: answer the scenario, not the keyword
In Red Hat Enterprise Linux shell scripting, a common mistake is to use `set -e` or conditional operators like `&&` when the goal is to capture the exit status regardless of success or failure. The correct approach is to assign `$?` unconditionally immediately after the command.
Trap categories for this question
Command / output trap
D is wrong because $? is evaluated before the command runs.
Detailed technical explanation
How to think about this question
The `$?` variable is a special shell parameter that always contains the exit status of the last foreground pipeline. In a script, capturing `$?` immediately after the command is critical because any subsequent command (like `echo` or a test) will overwrite it. A real-world scenario is a deployment script where a risky command (e.g., `curl` to a flaky API) must be logged regardless of success, and the script must continue to clean up or notify. Using `set -e` would abort the script, while `&&` would skip logging on failure—both violate the requirement.
KKey Concepts to Remember
- Read the scenario before looking for a memorised answer.
- Find the constraint that changes the correct option.
- Eliminate answers that are true in general but not in this case.
TExam Day Tips
- Watch for words such as best, first, most likely and least administrative effort.
- Review why wrong options are wrong, not only why the correct option is correct.
Key takeaway
Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Real-world example
How this comes up in practice
A practitioner preparing for the EX200 exam encounters this exact type of scenario on the job. The correct answer here is not the most general option — it is the best answer for the specific constraint described. Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option. Real exam questions reward reading the full scenario before eliminating options, because the constraint defines which answer fits.
What to study next
Got this wrong? Here's your next step.
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FAQ
Questions learners often ask
What does this EX200 question test?
Create simple shell scripts — This question tests Create simple shell scripts — Read the scenario before looking for a memorised answer..
What is the correct answer to this question?
The correct answer is: ./risky_command; rc=$?; echo $rc — Option C is correct because it runs the risky command, captures its exit status immediately after in the `$?` variable, and then echoes it for logging. The script continues regardless of the command's success or failure, which meets the requirement. The `$?` variable holds the exit status of the last executed foreground command, so assigning it to `rc` right after `./risky_command` ensures the correct value is stored.
What should I do if I get this EX200 question wrong?
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
What is the key concept behind this question?
Read the scenario before looking for a memorised answer.
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Last reviewed: Jul 4, 2026
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