Question

在Python中，我可以在uuid模块中始终生成seg错误。这可以通过从多个线程重复调用uuid.uuid1()来完成。经过一些挖掘，看来这个函数最终通过uuid_generate_time调用C ctypes函数：

来自uuid.py：

for libname in ['uuid', 'c']:
    try:
        lib = ctypes.CDLL(ctypes.util.find_library(libname))
    except:
        continue
    if hasattr(lib, 'uuid_generate_random'):
        _uuid_generate_random = lib.uuid_generate_random
    if hasattr(lib, 'uuid_generate_time'):
        _uuid_generate_time = lib.uuid_generate_time
        if _uuid_generate_random is not None:
            break  # found everything we were looking for

后来在uuid1()：

的定义中

def uuid1(node=None, clock_seq=None):
    """Generate a UUID from a host ID, sequence number, and the current time.
    If 'node' is not given, getnode() is used to obtain the hardware
    address.  If 'clock_seq' is given, it is used as the sequence number;
    otherwise a random 14-bit sequence number is chosen."""

    # When the system provides a version-1 UUID generator, use it (but don't
    # use UuidCreate here because its UUIDs don't conform to RFC 4122).
    if _uuid_generate_time and node is clock_seq is None:
        _buffer = ctypes.create_string_buffer(16)
        _uuid_generate_time(_buffer)
        return UUID(bytes=_buffer.raw)

我已经阅读了uuid_generate_time的手册页以及uuid.uuid1的Python文档，并且没有提到线程安全性。我认为这与它需要访问系统时钟和/或MAC地址这一事实有关，但这只是一个盲目猜测。

我想知道是否有人可以启发我？

以下是我用来生成seg错误的代码：

import threading, uuid

# XXX If I use a lock, I can avoid the seg fault
#uuid_lock = threading.Lock()
def test_uuid(test_func):
  for i in xrange(100):
    test_func()
    #with uuid_lock:
    #  test_func()

def test(test_func, threads):
  print 'Running %s with %s threads...' % (test_func.__name__, threads)

  workers = [threading.Thread(target=test_uuid, args=(test_func,)) for x in xrange(threads)]
  [x.start() for x in workers]
  [x.join() for x in workers]
  print 'Done!'

if __name__ == '__main__':
  test(uuid.uuid4, 8)
  test(uuid.uuid1, 8)

我得到的输出是：

Running uuid4 with 8 threads...
Done!
Running uuid1 with 8 threads...
Segmentation Fault (core dumped)

哦，我在Solaris上运行它......

Answer 1

文档没有说它的线程安全，所以你不能认为它是。它就这么简单。

查看uuid_generate_time的当前OpenIndiana source，导致段错误的原因并不完全明显。但是，虽然该函数确实使用了锁，但在执行许多初始化任务时它不会保持此锁。这可能与问题有关，但我无法指出竞争条件会导致故障的特定地点。您可以尝试在启动任何线程之前调用uuid1一次，看看是否会导致问题消失。虽然你最好只使用自己的锁，但因为不能保证Python代码本身是线程安全的。

uuid.uuid1，uuid_generate_time和threads

1 个答案: