bash command substitution on a script that executes a background task

Question

I'd like to return the results of a script that also kicks off a background task. The command substitution operator waits for the background task, making the call slow. I created the following example to illustrate the problem:

function answer {
    sleep 5 &
    echo string
}

echo $(answer)

Is there a way to call a command without waiting on any background jobs it creates?

Thanks,

Mark

score 3 · Accepted Answer · answered May 16 '14 at 15:44

3

The problem is that sleep inherits stdout and keeps it open. You can simply redirect stdout:

answer() {
  sleep 5 > /dev/null &
  echo "string"
}
echo "$(answer)"

answered May 16 '14 at 15:44

that other guy

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That's what was happening. Thank you!! – Mark Johnson May 16 '14 at 16:42

score 0 · Answer 2 · answered May 16 '14 at 15:31

0

If you are intending for the program to continue merrily along in the mean time while the function works, you can just call the function to run in the background.

function answer {
    sleep 5
    echo Second
}

echo $(answer) &
echo First

The output of which will be

First
Second

answered May 16 '14 at 15:31

Jonathan Wheeler

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Unfortunately, in the real-world example I need the contents of what the script returns. – Mark Johnson May 16 '14 at 15:35
I put these commands into a tmp.sh file, and then ran it using sh tmp.sh > out.txt and the output file worked just fine. – Jonathan Wheeler May 16 '14 at 15:43

bash command substitution on a script that executes a background task

2 Answers2

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