等待“许多过程中的1”完成

Question

Is there any built in feature in bash to wait for 1 out of many processes to finish? And then kill remaining processes?

pids=""
# Run five concurrent processes
for i in {1..5}; do
        ( longprocess ) &
        # store PID of process
        pids+=" $!"
done

if [ "one of them finished" ]; then
        kill_rest_of_them;
fi

I'm looking for "one of them finished" command. Is there any?

Answer 1

bash 4.3 added a -n flag to the built-in wait command, which causes the script to wait for the next child to complete. The -p option to jobs also means you may not need to store don't need to store the list of pics, as long as there aren't any background jobs that you don't want to wait on.

# Run five concurrent processes
for i in {1..5}; do
    ( longprocess ) &
done

wait -n
kill $(jobs -p)

Note that if there is another background job other than the 5 long processes that completes first, wait -n will exit when it completes. That would also mean you would still want to save the list of process ids to kill, rather than killing whatever jobs -p returns.

Answer 2

It's actually fairly easy:

#!/bin/bash
set -o monitor
killAll()
{
# code to kill all child processes
}

# call function to kill all children on SIGCHLD from the first one
trap killAll SIGCHLD

# start your child processes here

# now wait for them to finish
wait

You just have to be really careful in your script to use only bash built-in commands. You can't start any utilities that run as a separate process after you issue the trap command - any child process exiting will send SIGCHLD - and you can't tell where it came from.