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up vote 4 down vote favorite

Consider this:

tmux split-window -d program1
program2 # this program depends on some side effects produced by program1

In this case, program2 will start before program1 is ready. program1 will send signals/output when it has produced side-effects that can be consumed by program2.

My question: How can I write a shell script that will wait until program1 is ready before starting program2?

Observation: program2 does not consume program1 output, so this is not solved by connecting the processes with a pipe.

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Could you explain exactly how can the state of program1 (ready or not ready yet) be determined? And how exactly program1 sends those signals to program2? This can be useful because what you can do is make a script that looks for those signals, and only start program2 when those signals are detected. In any case if the circumstances permit it, it would be better if program2 itself deals with it. (for example. It goes to sleep until the signals wake it up) – Martín Canaval Jan 29 at 20:50
I just found the answer – Thiado de Arruda Jan 29 at 22:10

1 Answer

up vote 2 down vote

I found the solution here. In the end, it was not TMUX solution(initially I tought tmux would have some mechanism to syncronize the processes running in its panes)

The answer is using named pipes. Since program1 sends output when it is ready, all I need is something like this:

mkfifo /tmp/pipe
tmux split-window -d "program1 > /tmp/pipe 2>&1"
signal=`cat /tmp/pipe`
program2 # This will only run after program1 has output something
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