如果ThreadPoolExecutor中的一个任务可以引发异常，如何结束其中的所有任务

问题描述

我有一项任务要完成：

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

并且此任务中的逻辑可能引发异常

我有Execute方法和Executor：

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()

我希望如果至少有一个任务崩溃--不要等到其他任务完成后才停止一切。我如何才能正确执行此操作？

---

解决方案

若要在某个任务引发时放弃等待其他任务，可以使用带有FIRST_EXCEPTION标志的concurrent.futures.wait()：

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运行。没有办法强行阻止他们。