为什么 ThreadPoolExecutor 忽略异常？

Question

我有这样的代码：

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())

当我运行它时，我得到以下结果：

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

为什么第二个异常被忽略了？我想在每个线程上捕获并处理异常

Answer 1

因为异常由result() 引发并阻止正常的脚本流程。要查看第二个异常，您需要捕获第一个异常：

    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())