How to make timeout(1) behave in cron like it behaves in interactive shell

Running this command in an interactive bash:

$ timeout 1 sleep 2; echo $?
124

returns 124 after 1 second, as expected and as documented in timeout(1).

However, if I run the same as a cron job, or if I give that as a command string to bash, it does not:

$ bash -c "timeout 1 sleep 2; echo $?"
0

Adding -i to the bash invocation does not help, neither does using the --foreground parameter to timeout(1). I also tried the same with ksh and zsh, but always get the same result, so I guess it must be something inherent to the way timeout(1) works.

I searched the net a bit and found, that it could have to do with how the signalling proceeds through the process groups, but I could not find a solution how to make timeout work as expected in the non-interactive case.

Any hints on how I could achieve that? Ultimately what I want is to run a command in cron that is likely to block forever and I want to detect that case reliably.

You can try to run the command like this:

bash -c "timeout 1 sleep 2"; echo $?

Due to the double quotes, $? is being expanded before the bash command is invoked. It is being replaced by the exit status of the previous command (which was the first exit $?)

A quick demo

bash -c 'exit 42'
bash -c "timeout 1 sleep 2; echo $?"   # => 42

The solution is to use single quotes so that the current interactive bash process does not expand the variable

bash -c 'timeout 1 sleep 2; echo $?'   # => 124