How to limit `docker run` execution time? - docker

Question

I want to run a command inside a docker container. If the command takes more than 3 seconds to finish, the container should be deleted.
I thought I can achieve this goal by using --stop-timeout option in docker run.
But it looks like something goes wrong with my command.
For example, docker run -d --stop-timeout 3 ubuntu:14.04 sleep 100 command creates a docker container that lasts for more than 3 seconds. The container is not stopped or deleted after the 3rd second.
Do I misunderstand the meaning of --stop-timeout?
The document says
--stop-timeout Timeout (in seconds) to stop a container
Here's my docker version:
Client:
Version: 17.12.0-ce
API version: 1.35
Go version: go1.9.2
Git commit: c97c6d6
Built: Wed Dec 27 20:03:51 2017
OS/Arch: darwin/amd64
Server:
Engine:
Version: 17.12.0-ce
API version: 1.35 (minimum version 1.12)
Go version: go1.9.2
Git commit: c97c6d6
Built: Wed Dec 27 20:12:29 2017
OS/Arch: linux/amd64
Experimental: true
The API version is newer than 1.25.

Answer 1

You can try
timeout 3 docker run...
there is a PR on that subject
https://github.com/moby/moby/issues/1905
See also
Docker timeout for container?

Answer 2

The --stop-timeout option is the maximum amount of time docker should wait for your container to stop when using the docker stop command.
A container will stop when it's told to or when the command is running finishes, so if you change you sleep from 100 to 1, you'll see that the container is stopped after a second.
What I'll advice you to do is to change the ENTRYPOINT of your container to a script that you create, that will execute what you want and keep track of the execution time from within and exit when timeout.
After that you can start your container using the --rm option that will delete it once the script finishes.
A small example.
Dockerfile:
FROM ubuntu:16.04
ADD ./script.sh /script.sh
ENTRYPOINT /script.sh
script.sh:
#!/bin/bash
timeout=5
sleep_for=1
sleep 100 &
find_process=$(ps aux | grep -v "grep" | grep "sleep")
while [ ! -z "$find_process" ]; do
find_process=$(ps aux | grep -v "grep" | grep "sleep")
if [ "$timeout" -le "0" ]; then
echo "Timeout"
exit 1
fi
timeout=$(($timeout - $sleep_for))
sleep $sleep_for
done
exit 0
Run it using:
docker build -t testing .
docker run --rm testing
This script will execute sleep 100 in background, check if its still running and if the timeout of 5 seconds is reach then exit.
This might not be the best way to do it, but if you want to do something simple it may help.

Answer 3

docker run --rm ubuntu timeout 2 sh -c 'echo start && sleep 30 && echo finish'
will terminate after 2 seconds and finish will never be output

Answer 4

Depending on what exactly you want to achieve, the --ulimit parameter to docker run may do what you need. For example:
docker run --rm -it --ulimit cpu=1 debian:buster bash -c '(while true; do true; done)'
After about 1s, this will print Killed and return. With the --ulimit option, it would rune forever.
However, note that this only limits the CPU time, not the wall clock time. You can happily run sleep 24h with a --ulimit cpu=1 because sleep does not consume CPU time.

Answer 5

In my case, I had a docker container that started an Express server, and then remained running, and I wanted a simple test on CI to check that the container can start without any immediate error (such as configuration errors).
I made sure my code returned a non-zero exit code if something failed during start, and then ended up with this:
timeout 10 docker run [ container params ]; test $? -eq 124 && echo "Container ran for 10 seconds without failing"
This will send SIGTERM to the docker container after 10 seconds if it has not already died. If it's alive long enough for the timeout to occur, it will return 124, which is what the test is for. In other words, this verifies that the docker ran long enough to reach a timeout, any error (or early exit with code 0!) will be considered an error.