Why ThreadPoolExecutor ignoring exceptions?

Question

I have a code like this:

from concurrent.futures.thread import ThreadPoolExecutor
from time import sleep


def foo(message, time):
    sleep(time)
    print('Before exception in ' + message)
    raise Exception('OOOPS! EXCEPTION IN ' + message)
    
if __name__ == '__main__':
    executor = ThreadPoolExecutor(max_workers=2)
    future1 = executor.submit(foo, 'first', 1)
    future2 = executor.submit(foo, 'second', 3)
    print(future1.result())
    print(future2.result())

And when I run this, I get the following as a result:

Before exception in first
Traceback (most recent call last):
  ...
  File "...", line 8, in foo
    raise Exception('OOOPS! EXCEPTION IN ' + message)
Exception: OOOPS! EXCEPTION IN first
Before exception in second

Process finished with exit code 1

Why is the second exception being ignored? I want to catch and handle exceptions on each of the threads

score 1 · Accepted Answer · answered Nov 19 '20 at 09:55

1

Because exception is raised by result() and prevents normal script flow. To see the second exception you need to catch the first one:

    executor = ThreadPoolExecutor(max_workers=2)
    future1 = executor.submit(foo, 'first', 1)
    future2 = executor.submit(foo, 'second', 3)
    try:
        print(future1.result())
    except:
        pass
    print(future2.result())

answered Nov 19 '20 at 09:55

alex_noname

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And how can I stop a executor if an exception occurs in at least one of the threads – Pikachu Nov 19 '20 at 10:02
Perhaps it will help https://stackoverflow.com/questions/62474014/python-threadpoolexecutor-terminate-all-threads/62497987 – alex_noname Nov 19 '20 at 10:09

Why ThreadPoolExecutor ignoring exceptions?

1 Answers1