我打了一个非常简单的例子,希望它对你来说是一个很好的开始:
# this is important
from thread import start_new_thread, allocate_lock
# this is for show
from time import sleep
from random import randint
# global receiver
myresults = []
counter = 0
# lock that protects "myresults"
lock = allocate_lock()
#lock that protects "counter"
lock2 = allocate_lock()
# this does the url processing
def retrieve(url):
# sleep for 4-10s to simulate processing time
sleep(randint(4,10))
print "Done handling %s" % url
# thread safe retrieve
def retrieveLocking(url):
// global variables
global myresults, counter
// random processing time
sleep(randint(4,10))
print "[LOCKING] Done handling %s" % url
// request access to myresults' lock
lock.acquire()
myresults.append(url)
lock.release() // release lock!
// request access to counter's lock
lock2.acquire()
counter += 1
lock2.release() // release lock!
# here goes your main loop
start_new_thread(retrieve, ("A",))
start_new_thread(retrieve, ("B",))
start_new_thread(retrieve, ("C",))
# using locking
start_new_thread(retrieveLocking, ("A",))
start_new_thread(retrieveLocking, ("B",))
start_new_thread(retrieveLocking, ("C",))
在这种情况下,可以通过全局变量来处理查询结果。还有更复杂的方法可以从线程中检索数据。
如果您使用全局变量解决方案,我添加了锁。在并发编程中,当多个线程想要同时访问同一个资源时,可能会出现“竞争条件”。为了防止覆盖这些变量的旧状态,锁可以保护该资源。只有在锁被释放后,才允许下一个线程访问该资源。等待发布在这里为您处理:)
我希望这会有所帮助:)
// 编辑:
如果您以这种方式解决您的问题,那么实际上并没有办法确定所有线程是否都终止了。因此,如果您想坚持这个解决方案,请添加一个由每个线程增加的计数器(用锁保护计数器)并执行忙等待(无限循环,直到达到计数器)。这确实是不好的做法。但是,如果您想更精细地编写代码,事情会变得有点复杂,我假设您是并发编程的新手:
while True:
lock2.acquire()
if counter == totalthreads: # in this case 3
lock2.release()
break
lock2.release()
print "Done! %r" % myresults