某日，在排查线上问题时，在dump线程后发现了一些“诡异”的异常：
File "/usr/local/lib/python3.5/logging/__init__.py", line 1838, in info
root.info(msg, *args, **kwargs)
File "/usr/local/lib/python3.5/logging/__init__.py", line 1271, in info
Log 'msg % args' with severity 'INFO'.
File "/usr/local/lib/python3.5/logging/__init__.py", line 1279, in info
self._log(INFO, msg, args, **kwargs)
File "/usr/local/lib/python3.5/logging/__init__.py", line 1415, in _log
self.handle(record)
File "/usr/local/lib/python3.5/logging/__init__.py", line 1425, in handle
self.callHandlers(record)
File "/usr/local/lib/python3.5/logging/__init__.py", line 1487, in callHandlers
hdlr.handle(record)
File "/usr/local/lib/python3.5/logging/__init__.py", line 853, in handle
self.acquire()
File "/usr/local/lib/python3.5/logging/__init__.py", line 804, in acquire
self.lock.acquire()异常的表现是logging模块在打日志的时候卡在了获取锁这个地方，并且一直处于阻塞的状态。众所周知，Python的logging模块是线程安全的，因为logging模块在输出日志时，都要获取一把锁，而这把锁是一把可重入锁，对于不同的线程，在打日志前都要等待这把锁变为可用状态，才能持有这把锁并执行提交逻辑，典型的例子就像如下所示：
def handle(self, record):
"""
Conditionally emit the specified logging record.
Emission depends on filters which may have been added to the handler.
Wrap the actual emission of the record with acquisition/release of
the I/O thread lock. Returns whether the filter passed the record for
emission.
"""
rv = self.filter(record)
if rv:
self.acquire()
try:
self.emit(record)
finally:
self.release()
return rv由上可知，如果说一个线程在获取锁时，另一个线程迟迟没有释放锁，那么这个线程就会一直阻塞在获取锁的这个步骤，但实际上这种情况基本不可能出现，因为输出日志是一个十分快的过程，不需要耗费太多时间。
因为环境是在多进程+多线程环境下运行，所以有理由怀疑问题与并发时的竞态有关，于是先写一个多进程的例子：
import logging
import multiprocessing
import sys
from time import sleep
class CustomStreamHandler(logging.StreamHandler):
def emit(self, record):
sleep(0.1)
super(CustomStreamHandler, self).emit(record)
root = logging.getLogger()
root.setLevel(logging.DEBUG)
root.addHandler(CustomStreamHandler(sys.stdout))
def g():
logging.info(2)
logging.info(2)
def f():
logging.info(1)
logging.info(1)
p = multiprocessing.Process(target=f)
p.start()
g()主进程起了个新进程，调用f函数，打印日志“2”，接着并行调用g函数，打印日志“1”，为了看清楚问题，在logging handler获取锁后会sleep 0.1秒。
可以看到输出并没有问题：
1
2
1
2现在做一些修改，把主线程的g函数的调用修改为在主进程新建一个线程来执行：
import logging
import multiprocessing
import sys
from time import sleep
class CustomStreamHandler(logging.StreamHandler):
def emit(self, record):
sleep(0.1)
super(CustomStreamHandler, self).emit(record)
root = logging.getLogger()
root.setLevel(logging.DEBUG)
root.addHandler(CustomStreamHandler(sys.stdout))
def g():
logging.info(2)
logging.info(2)
def f():
logging.info(1)
logging.info(1)
import threading
t = threading.Thread(target=g)
p = multiprocessing.Process(target=f)
t.start()
p.start()执行输出，在程序输出两个2之后就立马阻塞住了：
2
2
阻塞...为什么两个相似的例子中会出现不一样的结果呢？
仔细观察，在这个例子中，线程先于进程开始调用，那么假设线程先执行，换而言之这时候root logger的锁便被线程池持有了，按照逻辑，线程马上进入睡眠0.1秒，而进程这个时候也想要输出日志，它也想要获取这把锁，却一直获取不到，甚至直到线程退出了进程也一直在阻塞着。
这一切表象背后的实际原因，实际上可以从上面推敲出，当进程被fork时，线程实际上已经持出这把锁了，所以说子进程在复制地址空间后它认为这把锁还在占用状态，于是就一直等着，但它不知道这把锁永远不会被释放了...
而为什么在前一个例子中却没有任何异常呢？事实上，再根据之前说的，这把锁时一把可重入锁（RLock），它是属于线程级别的锁，即是这把锁在哪个线程acquire就一定要在哪个线程被release，再来看个例子，虽然主进程中起了一个子进程，但实际上这两个进程所处于的线程都是一样的（主线程），而第二个例子中，f和g函数是处于完全不同的线程当中，就是f想要释放掉这把锁，它也无能为力，因为这把锁由g创建，只能被g释放，并且由于线程相异，g没有办法把锁释放。
如果做一个投机取巧，把锁替换掉会怎样呢：
import logging
import multiprocessing
import sys
from time import sleep
import threading
class CustomStreamHandler(logging.StreamHandler):
def emit(self, record):
sleep(0.1)
super(CustomStreamHandler, self).emit(record)
root = logging.getLogger()
root.setLevel(logging.DEBUG)
root.addHandler(CustomStreamHandler(sys.stdout))
def g():
print("g", threading.get_ident())
handler = logging.getLogger().handlers[]
logging.info(2)
logging.info(2)
print(id(handler))
print(id(handler.lock))
def f():
print("f", threading.get_ident())
handler = logging.getLogger().handlers[]
handler.createLock()
logging.info(1)
logging.info(1)
print(id(handler))
print(id(handler.lock))
import threading
print("main", threading.get_ident())
p = multiprocessing.Process(target=f)
t = threading.Thread(target=g)
t.start()
p.start()执行这段代码，输出：
main 140735977362240
g 123145405829120
f 140735977362240
2
1
2
1
4353914808
4353914808
4354039648
4349383520会发现不会有任何阻塞了，因为在子进程中直接把这把状态为locked的锁替换掉了（一把全新的RLock），这时候输出handler和lock的地址，发现两个函数中的handler地址都是同一个，而lock已经是两个不一样的了（Copy-On-Write）。
Solution
针对这个问题，Google很早之前已经给了一个解决方案：python-atfork
其主要思路是：对os.fork做一个monkeypatch，在fork调用时可触发三个自定义的hook：
def atfork(prepare=None, parent=None, child=None):- prepare：在fork子进程之前触发的函数。
- parent：在fork子进程之后父进程触发的函数。
- child：在fork子进程之后子进程触发的函数。
其中里面还给出了针对logging模块死锁问题的一个workaround：
def fix_logging_module():
logging = sys.modules.get('logging')
# Prevent fixing multiple times as that would cause a deadlock.
if logging and getattr(logging, 'fixed_for_atfork', None):
return
if logging:
warnings.warn('logging module already imported before fixup.')
import logging
if logging.getLogger().handlers:
# We could register each lock with atfork for these handlers but if
# these exist, other loggers or not yet added handlers could as well.
# Its safer to insist that this fix is applied before logging has been
# configured.
raise Error('logging handlers already registered.')
logging._acquireLock()
try:
def fork_safe_createLock(self):
self._orig_createLock()
atfork.atfork(self.lock.acquire,
self.lock.release, self.lock.release)
# Fix the logging.Handler lock (a major source of deadlocks).
logging.Handler._orig_createLock = logging.Handler.createLock
logging.Handler.createLock = fork_safe_createLock
# Fix the module level lock.
atfork.atfork(logging._acquireLock,
logging._releaseLock, logging._releaseLock)
logging.fixed_for_atfork = True
finally:
logging._releaseLock()确保了在fork子进程之后，锁是处于release状态的。
而在Python 3.7版本之后，官方也给出了标准库的方法实现相同的功能：
os.register_at_fork(*, before=None, after_in_parent=None, after_in_child=None)
Register callables to be executed when a new child process is forked using os.fork() or similar process cloning APIs. The parameters are optional and keyword-only. Each specifies a different call point.
- before is a function called before forking a child process.
- after_in_parent is a function called from the parent process after forking a child process.
- after_in_child is a function called from the child process.
