Bash : run a command and wait if necessary (contrary of timeout)

Question

I have a 10 second interval. I want to run a command in a shell, then wait to complete the 10 sec interval. If the commands takes more than 10 seconds, its OK.

I am looking for an equivalent to "foo" function below :

>> foo 10 python myscript.py

If myscript.py takes 2 seconds, the shell should wait 8 seconds afterwhile.
If myscript.py takes 5 seconds, the shell should wait 5 seconds afterwhile.
If myscript.py takes 12 seconds, the shell should not wait. It should not timeout myscript.py either !

Does this "foo" function exist ?

I wrote a answer for a similar problem over here: http://stackoverflow.com/a/41108021/771740 — JonatasTeixeira, Dec 12 '16 at 19:36
I answer a similar question over here: http://stackoverflow.com/a/41108021/771740 — JonatasTeixeira, Dec 13 '16 at 21:43

score 0 · Accepted Answer · answered Jul 24 '16 at 15:16

Just run both sleep 10 and your script in the background, then wait for both to complete.

python myscript.py &
sleep 10 &
wait

You can define a shell function to wrap simple commands (a command name and its arguments; no pipes, redirections, && and ||, etc.)

foo () {
    t=$1
    shift
    "$@" &
    sleep "$t" &
    wait
}

foo 10 python myscript.py

Bash : run a command and wait if necessary (contrary of timeout)

1 Answers1