生产者使用者3个线程，每个在python中

Question

I'm trying to do a producer consumer program. I got it working just fine with one thread for each and I'm trying to modify it to run three threads of each. It appears that each of the consumer threads is trying to consume each released item.

# N is the number of slots in the buffer
N = 8
n = 0
i=0
j=0
# initialise buf with the right length, but without values
buf = N * [None]


free = threading.Semaphore(N)
items = threading.Semaphore(0)
block = threading.Semaphore(1)

# a function for the producer thread
def prod(n, j):
    while True:
        time.sleep(random.random())
        free.acquire()
        # produce a number and add it to the buffer
        buf[i] = n
        #print("produced")
        j = (j + 1) % N
        n += 1
        items.release()


# a function for the consumer thread
def cons(th):
    global i
    while True:
        time.sleep(random.random())
        #acquire items to allow the consumer to print. 
        items.acquire()
        print(buf[i])
        print("consumed, th:{} i:{}".format(th, i))
        i = (i + 1) % N
        #time.sleep(3)
        free.release()





# a main function
def main():
    p1 = threading.Thread(target=prod, args=[n,j])
    p2 = threading.Thread(target=prod, args=[n,j])
    p3 = threading.Thread(target=prod, args=[n,j])
    c1 = threading.Thread(target=cons, args=[1])
    c2 = threading.Thread(target=cons, args=[2])
    c3 = threading.Thread(target=cons, args=[3])

    p1.start()
    p2.start()
    p3.start()
    c1.start()
    c2.start()
    c3.start()
    p1.join()
    p2.join()
    p3.join()
    c1.join()
    c2.join()
    c3.join()


main()

Any help is appreciated. I'm really at a loss with this one.

Answer 1

When your code in a thread acquires a semaphore, it should then subsequently release the same semaphore. So instead of:

items.acquire()
...
free.release()

your code must do, eg

items.acquire()
...
items.release()