以下の例では、プログラムを複数回実行すると、毎回新しいIDで新しいスレッドが生成されます。1.タスクの完了時にすべてのスレッドを終了するにはどうすればよいですか?2.スレッドに名前/IDを割り当てるにはどうすればよいですか?
import threading, Queue
THREAD_LIMIT = 3
jobs = Queue.Queue(5) # This sets up the queue object to use 5 slots
singlelock = threading.Lock() # This is a lock so threads don't print trough each other
# list
inputlist_Values = [ (5,5),(10,4),(78,5),(87,2),(65,4),(10,10),(65,2),(88,95),(44,55),(33,3) ]
def DoWork(inputlist):
print "Inputlist received..."
print inputlist
# Spawn the threads
print "Spawning the {0} threads.".format(THREAD_LIMIT)
for x in xrange(THREAD_LIMIT):
print "Thread {0} started.".format(x)
# This is the thread class that we instantiate.
worker().start()
# Put stuff in queue
print "Putting stuff in queue"
for i in inputlist:
# Block if queue is full, and wait 5 seconds. After 5s raise Queue Full error.
try:
jobs.put(i, block=True, timeout=5)
except:
singlelock.acquire()
print "The queue is full !"
singlelock.release()
# Wait for the threads to finish
singlelock.acquire() # Acquire the lock so we can print
print "Waiting for threads to finish."
singlelock.release() # Release the lock
jobs.join() # This command waits for all threads to finish.
class worker(threading.Thread):
def run(self):
# run forever
while 1:
# Try and get a job out of the queue
try:
job = jobs.get(True,1)
singlelock.acquire() # Acquire the lock
print self
print "Multiplication of {0} with {1} gives {2}".format(job[0],job[1],(job[0]*job[1]))
singlelock.release() # Release the lock
# Let the queue know the job is finished.
jobs.task_done()
except:
break # No more jobs in the queue
def main():
DoWork(inputlist_Values)