在Python中,我可以在
来自uuid.py:
我已经阅读了
我得到的输出是:
uuid
模块中一致地生成一个段错误。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,似乎该函数最终通过ctypes
调用C uuid_generate_time
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
接着在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
的手册以及Python文档中关于 uuid.uuid1
的部分,但是没有提到线程安全的问题。我猜测这可能与需要访问系统时钟和/或MAC地址有关,但这只是一个盲猜。
我想知道是否有人能为我解答?
以下是我用来生成段错误的代码:
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,我正在Solaris上运行这个程序...