如果其中一个可以引发异常，如何结束 ThreadPoolExecutor 中的所有任务

Question

我有一个任务要做：

def task(body):
    # some logic that which can throw an exception
    # if something goes wrong
    do_task(body) 

而且这个任务里面的逻辑可以抛出异常

我有执行器的执行方法：

def execute():
    executor = ThreadPoolExecutor(max_workers=4)
    future1 = executor.submit(task, body1)
    future2 = executor.submit(task, body2)
    future3 = executor.submit(task, body3)
    future4 = executor.submit(task, body4)
    
    result1 = future1.result()
    result2 = future2.result()
    result3 = future3.result()
    result4 = future4.result()

如果至少有一项任务崩溃，我希望 - 不要等待其他任务完成并停止一切。我怎样才能正确地做到这一点？

Answer 1

要在一个加注时放弃等待其他任务，您可以使用 concurrent.futures.wait() 和 FIRST_EXCEPTION 标志：

def execute():
    executor = ThreadPoolExecutor(max_workers=4)
    future1 = executor.submit(task, body1)
    future2 = executor.submit(task, body2)
    future3 = executor.submit(task, body3)
    future4 = executor.submit(task, body4)

    done, not_done = concurrent.futures.wait(
        [future1, future2, future3, future4],
        return_when=concurrent.futures.FIRST_EXCEPTION
    )
    if not_done:
        # at least one future has raised - you can return here
        # or propagate the exception
        #list(not_done)[0].result()  # re-raises exception here
        return  # ignores exception and returns

    result1 = future1.result()
    result2 = future2.result()
    result3 = future3.result()
    result4 = future4.result()
    ...

请注意，其余任务仍将在后台继续运行。没有办法强行阻止。