为什么信号处理程序中的打印操作可能会改变死锁情况？ [英] Why print operation within signal handler may change deadlock situation?

问题描述

我得到了一个简单的程序，如下所示：

I got simple program as below:

import threading
import time
import signal

WITH_DEADLOCK = 0

lock = threading.Lock()

def interruptHandler(signo, frame):
    print str(frame), 'received', signo
    lock.acquire()
    try:
        time.sleep(3)
    finally:
        if WITH_DEADLOCK:
            print str(frame), 'release'
        lock.release()

signal.signal(signal.SIGINT, interruptHandler)
for x in xrange(60):
    print time.strftime("%H:%M:%S"), 'main thread is working'
    time.sleep(1)

因此，如果启动该程序，甚至在3秒内按两次Ctrl + C，也不会出现死锁。每次按Ctrl + C，都会显示正确的行。
如果您更改WITH_DEADLOCK = 1并按Ctrl + C两次（3秒内），程序将被挂起。

So, if you start that program and even Ctrl+C is pressed twice within 3 seconds, there is no deadlock. Each time you press Ctrl + C proper line is displayed. If you change WITH_DEADLOCK=1 and you would press Ctrl+C twice (withing 3 seconds) then program will be hung.

有人可以解释为什么打印吗

Does anybody may explain why print operation make such difference?

（我的python版本是2.6.5）

(My python version is 2.6.5)

推荐答案

说实话，我认为JF Sebastian的评论在这里是最合适的答案-您需要使信号处理程序可重入，但目前尚不可以重入，而且最令人惊讶的是，如果没有print语句，它仍然可以工作。

To be honest I think J.F. Sebastian's comment is the most appropriate answer here - you need to make your signal handler reentrant, which it currently isn't, and it is mostly just surprising that it works anyway without the print statement.

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