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I think I am doing something wrong when I tried to implement simple barrier in linux shell script but seem to fail spectacularly.


The problem is I can see <snippet1> and all instances of <script1> is launched
But output of code block in <snippet2> appearing before the output of instances of <script1>.
Because there is a dependency where <script1> can not run after <snippet2> is executed it runs into all sort of errors.

Try replacing your barrier with this:

This always going to be larger than 0, because grep itself is included in ps output. (btw the flag "f" makes no sense here.)

This trick I've learned from an old UNIX guru:
This will list all processes containing "something" except the grep The reason is, the command will be "grep [s]omething", however [] is replaced by the shell, and "grep something" will actually be executed, that doesn't match "[s]omething".

Btw, "wait" is a much better solution, I agree.

Cheers,
bzt

Not necessarily (and that can lead to some fun bugs): The shell will spawn all parts of the pipeline in parallel. So whether "grep" has been exec'd by the time ps gets to that process is not defined. There is no sequence that makes that a necessity, and indeed, in some situations ps fails to output the line with "grep" in it (if grep happens to get a very small PID and ps has already iterated over it by the time grep is exec'd). And so you end up with the grep line sometimes being in the output and sometimes not. Happy debugging!

The trick you showed is a good one, however. I already knew it, but it is a good one for newbies (regular expressions don't have to match themselves). Alternatively, you can filter the "grep" line afterwards with "grep -v grep". But for this particular problem, Octocontrabass was entirely correct: There is no need to iterate over all processes when you only need to wait for your child processes.

thanks I used wait and apparently working much better.