Will "hello" always be printed?

[Cameron Simpson]

On 07Oct2022 20:16, Robin van der veer <robinvdveer at gmail.com> wrote:
>If I have two processes communicating through a JoinableQueue, and I do the
>following:
>
>process 1:
>
>    queue.put(1) #unfished tasks = 1
>    queue.join() #block until unfished tasks = 0
>    print('hello')[/python]
>
>process 2:
>
>    queue.get()
>    queue.task_done() #unfished tasks = 0
>    queue.put(1) #unfinished tasks 1[/python]
>the unfished tasks refers to what is written in the documentation (
>https://docs.python.org/3/library/multiprocessing.html#multiprocessing.JoinableQueue.join
>)
>
>will 'hello' always be printed? Or is there a chance that the put in
>process 2 executes before process 1 noticed that it should unblock?

I had to read this closely. Yes, the second `put(1)` could execute 
before the `join()` commences (or tests), and the `hello` would be 
blocked still.

>It seems that the whole point of join() is that 'hello' should always be
>printed, but I just want to make sure that I understand it correctly.

That's the purpose of using `join`, but you need to use it correctly.  
The "some tasks are not completed" condition which `join` supports 
doesn't fit what you're doing.

So yes, you're correct in your concern.

Maybe 2 queues would suit you better? Maybe not if they are common.

Maybe some kind of blocking counter, so that you could track task 
counts? You'd need to make one, but something which allowed:

     count = queue.put(1)
     queue.wait_for(count)
     print('hello')

and at the other end:

     queue.get()
     queue.task_done()   # bumps the counter
     count2 = queue.put(1)

Here, the counter would not be the internal "unfinished tasks" counter 
but instead a distinct counter which always went up. Probably a pair: 
tasks submitted by `put` and tasks completed by `task_done`. You could 
subclass `JoinableQueue` and add implementations of these counters and 
add a `wait_for(count)` method.

This amounts to assigning each "task" a unique id (the counter) and a 
means to wait for that id.

Cheers,
Cameron Simpson <cs at cskk.id.au>