python - BlockingIOError 异常被忽略；我应该担心吗？

Question

我正在编写一个包含 20,000 个任务的脚本，每个任务进行一个子进程调用和一个或两个 TCP 调用。为了使这不会花费一整天，我使用了 Python 的新 asyncio。

但是，我担心脚本运行时 Python 会打印出这些错误:

Exception ignored when trying to write to the signal wakeup fd:
BlockingIOError: [Errno 11] Resource temporarily unavailable

它将打印一堆，但不会引发任何异常。我确实得到了关于太多打开的文件的OSError并且之前与服务器断开了连接，但是我使用信号量一次只允许100个连接到每个服务器并且只有700个连接总数。

由于 Python 没有引发任何异常，我无法捕获错误。但它似乎不会影响脚本。

我应该担心这些错误吗？如果是这样，我需要做什么来摆脱它们？如果没有，我该如何摆脱它们，使它们不在我的程序输出中？

此外，如果这些错误很严重，为什么 Python 会忽略它们而不是引发异常？

Answer 1

看起来限制因素是运行大量短命的 subprocesses。来自Python bug tracker :

"Exception ignored when trying to write to the signal wakeup fd" message comes from the signal handler in Modules/signalmodule.c. The problem is that Python gets a lot of SIGCHLD signals (the test scripts creates +300 processes per second on my computer). The producer (signal handler writing the signal number into the "self" pipe) is faster than the consumer (BaseSelectorEventLoop._read_from_self callback).

With the patch, I start getting messages with 140 concurrent processes, which is much better :-) IMO more than 100 concurrent processes is crazy, don't do that at home :-) I mean processes with a very short lifetime. The limit is the number of SIGCHLD per second, so the number of processes which end at the same second.

我更改了我的代码以限制多少 create_subprocess_exec 可以同时运行。当我低于 35 时，我不再看到错误，尽管我可能会将其设置为 20 以确保万无一失。您的里程可能会有所不同。

async def myTask(stuff, semaphore, loop):
    with semaphore:
        process = await asyncio.create_subprocess_exec('short_program', loop=loop)

def taskRunner(stuffs):
    loop = asyncio.get_event_loop()
    semaphore = asyncio.Semaphore(20)  # limit how many can run at a time
    tasks = [
        asyncio.ensure_future(myTask(semaphore, loop))
        for i in range(20000)
    ]
    loop.run_until_complete(asyncio.gather(*tasks))
    loop.close()