This answer to Command line command to auto-kill a command after a certain amount of time
proposes a 1-line method to timeout a long-running command from the bash command line:
( /path/to/slow command with options ) & sleep 5 ; kill $!
But it's possible that a given "long-running" command may finish earlier than the timeout.
(Let's call it a "typically-long-running-but-sometimes-fast" command, or tlrbsf for fun.)
So this nifty 1-liner approach has a couple of problems.
First, the sleep isn't conditional, so that sets an undesirable lower bound on the time taken for the sequence to finish. Consider 30s or 2m or even 5m for the sleep, when the tlrbsf command finishes in 2 seconds — highly undesirable.
Second, the kill is unconditional, so this sequence will attempt to kill a non-running process and whine about it.
So...
Is there a way to timeout a typically-long-running-but-sometimes-fast ("tlrbsf") command that
... and, for bonus points
such that the stdin/stdout/stderr of the tlrbsf command can be redirected, same as if it had been run directly?
If so, please share your code. If not, please explain why.
I have spent awhile trying to hack the aforementioned example but I'm hitting the limit of my bash skills.
I think this is precisely what you are asking for:
http://www.bashcookbook.com/bashinfo/source/bash-4.0/examples/scripts/timeout3
#!/bin/bash
#
# The Bash shell script executes a command with a time-out.
# Upon time-out expiration SIGTERM (15) is sent to the process. If the signal
# is blocked, then the subsequent SIGKILL (9) terminates it.
#
# Based on the Bash documentation example.
# Hello Chet,
# please find attached a "little easier" :-) to comprehend
# time-out example. If you find it suitable, feel free to include
# anywhere: the very same logic as in the original examples/scripts, a
# little more transparent implementation to my taste.
#
# Dmitry V Golovashkin <[email protected]>
scriptName="${0##*/}"
declare -i DEFAULT_TIMEOUT=9
declare -i DEFAULT_INTERVAL=1
declare -i DEFAULT_DELAY=1
# Timeout.
declare -i timeout=DEFAULT_TIMEOUT
# Interval between checks if the process is still alive.
declare -i interval=DEFAULT_INTERVAL
# Delay between posting the SIGTERM signal and destroying the process by SIGKILL.
declare -i delay=DEFAULT_DELAY
function printUsage() {
cat <<EOF
Synopsis
$scriptName [-t timeout] [-i interval] [-d delay] command
Execute a command with a time-out.
Upon time-out expiration SIGTERM (15) is sent to the process. If SIGTERM
signal is blocked, then the subsequent SIGKILL (9) terminates it.
-t timeout
Number of seconds to wait for command completion.
Default value: $DEFAULT_TIMEOUT seconds.
-i interval
Interval between checks if the process is still alive.
Positive integer, default value: $DEFAULT_INTERVAL seconds.
-d delay
Delay between posting the SIGTERM signal and destroying the
process by SIGKILL. Default value: $DEFAULT_DELAY seconds.
As of today, Bash does not support floating point arithmetic (sleep does),
therefore all delay/time values must be integers.
EOF
}
# Options.
while getopts ":t:i:d:" option; do
case "$option" in
t) timeout=$OPTARG ;;
i) interval=$OPTARG ;;
d) delay=$OPTARG ;;
*) printUsage; exit 1 ;;
esac
done
shift $((OPTIND - 1))
# $# should be at least 1 (the command to execute), however it may be strictly
# greater than 1 if the command itself has options.
if (($# == 0 || interval <= 0)); then
printUsage
exit 1
fi
# kill -0 pid Exit code indicates if a signal may be sent to $pid process.
(
((t = timeout))
while ((t > 0)); do
sleep $interval
kill -0 $$ || exit 0
((t -= interval))
done
# Be nice, post SIGTERM first.
# The 'exit 0' below will be executed if any preceeding command fails.
kill -s SIGTERM $$ && kill -0 $$ || exit 0
sleep $delay
kill -s SIGKILL $$
) 2> /dev/null &
exec "$@"