Parallel execution patterns in Bash Scripting - Practice Problems & Coding Challenges

Challenge - 5 Problems

💻 Command Output

intermediate

Output of background jobs with wait

What will be the output of this script when run in a bash shell?

Bash Scripting

echo "Start"
(sleep 1; echo "Job 1 done") &
(sleep 2; echo "Job 2 done") &
wait
echo "All jobs finished"

A Start\nAll jobs finished\nJob 1 done\nJob 2 done

B Start\nJob 1 done\nJob 2 done\nAll jobs finished

C Job 1 done\nJob 2 done\nStart\nAll jobs finished

D Start\nJob 2 done\nJob 1 done\nAll jobs finished

💡 Hint

Remember that 'wait' pauses the script until all background jobs finish.

The script prints 'Start' immediately, then runs two background jobs that sleep and print messages. The 'wait' command pauses the script until both jobs finish, so their outputs appear before 'All jobs finished'. Job 1 finishes before Job 2 due to shorter sleep.

📝 Syntax

intermediate

Correct syntax for running commands in parallel

Which option correctly runs three commands in parallel and waits for all to finish?

A cmd1 & cmd2 & cmd3 & wait &

B cmd1 & cmd2 & cmd3 wait

C cmd1 & cmd2 & cmd3 & wait

D cmd1 cmd2 cmd3 & wait

💡 Hint

Use '&' to run commands in background and 'wait' to pause until all finish.

Option C correctly runs cmd1, cmd2, and cmd3 in background using '&' and then waits for all to finish with 'wait'. Option C misses '&' before cmd3 so cmd3 runs in foreground. Option C runs wait in background which is incorrect. Option C runs all commands sequentially in foreground.

🔧 Debug

advanced

Fix the issue with parallel job output mixing

This script runs two background jobs but their outputs sometimes mix on the terminal. What is the main cause?

Bash Scripting

echo "Start"
(sleep 1; echo "Job 1 done") &
(sleep 1; echo "Job 2 done") &
wait
echo "End"

A The sleep times are different causing output order issues.

B The 'wait' command is missing causing jobs to overlap.

C The echo commands are missing quotes causing syntax errors.

D Both jobs write to stdout simultaneously causing output mixing.

💡 Hint

Think about how multiple background jobs write to the terminal at the same time.

When multiple background jobs write to the terminal at the same time, their outputs can interleave or mix because stdout is shared. This is normal behavior unless output is redirected or synchronized.

🚀 Application

advanced

Limit number of parallel jobs in a loop

You want to run 10 commands in parallel but only allow 3 to run at the same time. Which approach achieves this?

Bash Scripting

for i in {1..10}; do
  some_command $i &
  # limit to 3 parallel jobs
  # ???
done
wait

A Use a counter and 'wait' every 3 jobs: if (( i % 3 == 0 )); then wait; fi

B Run all commands in background without wait to limit jobs

C Use 'sleep 3' inside the loop to delay jobs

D Use 'wait' only after the loop to limit jobs

💡 Hint

You can pause the loop every few jobs to limit concurrency.

Option A uses a counter to wait after every 3 jobs, ensuring only 3 run in parallel at once. Option A runs all jobs at once, no limit. Option A delays but does not limit concurrency. Option A waits only at the end, no limit during loop.

🧠 Conceptual

expert

Understanding job control and process IDs in parallel scripts

In a bash script, you run multiple background jobs and want to track their process IDs (PIDs) to wait for specific ones. Which method correctly captures and waits for each job's PID?

A

cmd1 &
pid1=$!
cmd2 &
pid2=$!
wait $pid1
wait $pid2

B

cmd1 & pid1=$!
cmd2 & pid2=$!
wait $pid1
wait $pid2

C

pid1=$!; cmd1
pid2=$!; cmd2
wait $pid1
wait $pid2

D

pid1=$!; cmd1 &
pid2=$!; cmd2 &
wait $pid1
wait $pid2

💡 Hint

The special variable $! holds the PID of the last background job started.

Option A correctly runs cmd1 and cmd2 in background, then immediately captures their PIDs with $!. Option A captures $! before starting jobs, so PIDs are wrong. Option A runs commands in foreground, so $! is empty or unrelated.