Ujo*_*oux 5 python ctypes multiprocessing
我正在启动3个进程,我希望它们在与进程(i)对应的索引处将字符串放入共享数组中.
看下面的代码,生成的输出是:
['test 0', None, None]
['test 1', 'test 1', None]
['test 2', 'test 2', 'test 2']
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为什么"测试0"得到通过覆盖test 1,并test 1通过test 2?
我想要的是(顺序并不重要):
['test 0', None, None]
['test 0', 'test 1', None]
['test 0', 'test 1', 'test 2']
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代码 :
#!/usr/bin/env python
import multiprocessing
from multiprocessing import Value, Lock, Process, Array
import ctypes
from ctypes import c_int, c_char_p
class Consumer(multiprocessing.Process):
def __init__(self, task_queue, result_queue, arr, lock):
multiprocessing.Process.__init__(self)
self.task_queue = task_queue
self.result_queue = result_queue
self.arr = arr
self.lock = lock
def run(self):
proc_name = self.name
while True:
next_task = self.task_queue.get()
if next_task is None:
self.task_queue.task_done()
break
answer = next_task(arr=self.arr, lock=self.lock)
self.task_queue.task_done()
self.result_queue.put(answer)
return
class Task(object):
def __init__(self, i):
self.i = i
def __call__(self, arr=None, lock=None):
with lock:
arr[self.i] = "test %d" % self.i
print arr[:]
def __str__(self):
return 'ARC'
def run(self):
print 'IN'
if __name__ == '__main__':
tasks = multiprocessing.JoinableQueue()
results = multiprocessing.Queue()
arr = Array(ctypes.c_char_p, 3)
lock = multiprocessing.Lock()
num_consumers = multiprocessing.cpu_count() * 2
consumers = [Consumer(tasks, results, arr, lock) for i in xrange(num_consumers)]
for w in consumers:
w.start()
for i in xrange(3):
tasks.put(Task(i))
for i in xrange(num_consumers):
tasks.put(None)
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我正在运行Python 2.7.3(Ubuntu)
这个问题看起来与此类似.在那里,JF Sebastian推测arr[i]指向arr[i]一个内存地址的赋值只对进行赋值的子进程有意义.其他子进程在查看该地址时检索垃圾.
至少有两种方法可以避免这个问题.一个是使用multiprocessing.manager列表:
import multiprocessing as mp
class Consumer(mp.Process):
def __init__(self, task_queue, result_queue, lock, lst):
mp.Process.__init__(self)
self.task_queue = task_queue
self.result_queue = result_queue
self.lock = lock
self.lst = lst
def run(self):
proc_name = self.name
while True:
next_task = self.task_queue.get()
if next_task is None:
self.task_queue.task_done()
break
answer = next_task(lock = self.lock, lst = self.lst)
self.task_queue.task_done()
self.result_queue.put(answer)
return
class Task(object):
def __init__(self, i):
self.i = i
def __call__(self, lock, lst):
with lock:
lst[self.i] = "test {}".format(self.i)
print([lst[i] for i in range(3)])
if __name__ == '__main__':
tasks = mp.JoinableQueue()
results = mp.Queue()
manager = mp.Manager()
lst = manager.list(['']*3)
lock = mp.Lock()
num_consumers = mp.cpu_count() * 2
consumers = [Consumer(tasks, results, lock, lst) for i in xrange(num_consumers)]
for w in consumers:
w.start()
for i in xrange(3):
tasks.put(Task(i))
for i in xrange(num_consumers):
tasks.put(None)
tasks.join()
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另一种方法是使用具有固定大小的共享数组,例如mp.Array('c', 10).
import multiprocessing as mp
class Consumer(mp.Process):
def __init__(self, task_queue, result_queue, arr, lock):
mp.Process.__init__(self)
self.task_queue = task_queue
self.result_queue = result_queue
self.arr = arr
self.lock = lock
def run(self):
proc_name = self.name
while True:
next_task = self.task_queue.get()
if next_task is None:
self.task_queue.task_done()
break
answer = next_task(arr = self.arr, lock = self.lock)
self.task_queue.task_done()
self.result_queue.put(answer)
return
class Task(object):
def __init__(self, i):
self.i = i
def __call__(self, arr, lock):
with lock:
arr[self.i].value = "test {}".format(self.i)
print([a.value for a in arr])
if __name__ == '__main__':
tasks = mp.JoinableQueue()
results = mp.Queue()
arr = [mp.Array('c', 10) for i in range(3)]
lock = mp.Lock()
num_consumers = mp.cpu_count() * 2
consumers = [Consumer(tasks, results, arr, lock) for i in xrange(num_consumers)]
for w in consumers:
w.start()
for i in xrange(3):
tasks.put(Task(i))
for i in xrange(num_consumers):
tasks.put(None)
tasks.join()
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我推测,当mp.Array(ctypes.c_char_p, 3)没有这个时,它的工作原因是因为它mp.Array('c', 10)有一个固定的大小,所以内存地址永远不会改变,同时mp.Array(ctypes.c_char_p, 3)有一个可变的大小,所以当arr[i]分配给一个更大的字符串时,内存地址可能会改变.
也许这就是文档警告的时候,
虽然可以将指针存储在共享内存中,但请记住这将引用特定进程的地址空间中的位置.但是,指针很可能在第二个进程的上下文中无效,并且尝试从第二个进程取消引用指针可能会导致崩溃.
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