Question

I am playing with using flock, a bash command for file locks to prevent two different instances of the code from running more than once.

I am using this testing code:

( ( flock -x 200 ;  sleep 10 ; echo "original finished" ; ) 200>./test.lock ) &
( sleep 2 ; ( flock -x -w 2 200 ; echo "a finished" ) 200>./test.lock ) &

I am running 2 subshells (backgrounded). The (flock NUM; ...) NUM>FILE syntax is from flock‘s man page.

I expect that the first subshell will get an exclusive lock on test.lock, then wait 10 seconds, then print “original finished”, all the time holding the lock. The second subshell will start at more or less the same time, wait 2 seconds, then try to get a lock on test.lock, but timeout after 2 seconds. If it gets a lock, then it’ll print “a finished”. If it doesn’t get the lock, that subshell should stop, and nothing should be printed.

Since the first subshell is waiting longer, it will keep the lock for 10 seconds, so the second subshell should not get the lock, and shouldn’t finish. i.e. one should see “original finished” printed and not both.

What actually happens is that “a finished” is printed, then “original finished” is printed.

This implies that that the second subshell is either (a) not using the same lock as the first subshell or (b) that it fails to get the lock, but continues to execute or (c) something else.

Why don’t those locks work as I expect?

Answer 1

The issue is that, if the flock process fails to get the lock within the timeout, it has no way of killing the parent process (i.e. the shell that spawned it) – all it can do is return a failure return code. You need to check that return code before continuing:

flock <params> && <do other stuff>

so

( ( flock -x 200 ;  sleep 10 ; echo "original finished" ; ) 200>./test.lock ) & ( sleep 2 ; ( flock -x -w 2 200 && echo "a finished" ) 200>./test.lock ) &

does what you want.