我是python的新手，我有以下简单代码：
def f(ID, N):
# global lock
if ID == 2:
time.sleep(0.5)
if ID == 1:
lock.acquire()
for i in range(5):
print(f'{N}: {time.ctime(time.time())} \n')
print(f"{i}: lock.locked() = {lock.locked()} \n")
time.sleep(1)
if i == 2 and ID == 1:
lock.release()
t = time.time()
class myThread2 (threading.Thread):
def __init__(self, threadID, threadname):
threading.Thread.__init__(self)
self.threadID = threadID
self.name = threadname
def run(self):
f(self.threadID, self.name)
print(f'{self.name} finished')
lock = threading.Lock()
# Create new threads objects
thread1 = myThread2(1, "Thread-1")
thread2 = myThread2(2, "Thread-2")
# Start new Threads
thread1.start()
thread2.start()
for i in range(5):
print(f'\t\t\t main programm: {time.ctime(time.time())} \n')
time.sleep(1)
# thread1.join()
# thread2.join()
print(time.time()-t)
输出为：
Thread-1: Thu Sep 27 15:04:29 2018
main programm: Thu Sep 27 15:04:29 2018
0: lock.locked() = True
Thread-2: Thu Sep 27 15:04:29 2018
0: lock.locked() = True
main programm: Thu Sep 27 15:04:30 2018
Thread-1: Thu Sep 27 15:04:30 2018
1: lock.locked() = True
Thread-2: Thu Sep 27 15:04:31 2018
1: lock.locked() = True
main programm: Thu Sep 27 15:04:31 2018
Thread-1: Thu Sep 27 15:04:31 2018
2: lock.locked() = True
Thread-2: Thu Sep 27 15:04:32 2018
2: lock.locked() = True
main programm: Thu Sep 27 15:04:32 2018
Thread-1: Thu Sep 27 15:04:32 2018
3: lock.locked() = False
Thread-2: Thu Sep 27 15:04:33 2018
3: lock.locked() = False
main programm: Thu Sep 27 15:04:33 2018
Thread-1: Thu Sep 27 15:04:33 2018
4: lock.locked() = False
Thread-2: Thu Sep 27 15:04:34 2018
4: lock.locked() = False
5.130102634429932Thread-1 finished
>>> Thread-2 finished
所以我的问题如下：为什么lock.acquire（）不能阻止线程2被执行直到i = 2和f函数中的lock.release（）命令执行？我希望它为线程1执行前3个循环，并且仅在该线程2进入for循环之后才执行。但是它们thread2在time.sleep（0.5）之后立即进入for循环。这是为什么？在输出中，thread1使for循环中断后，lock.locked（）= True，但它对thread2没有任何影响。错误在哪里，您能帮我吗？
答案 0 :(得分：0)
import time
from threading import Thread, Lock, current_thread
lock = Lock()
def fu():
lock.acquire()
for i in range(5):
print('Hello from ' + current_thread().name)
if i == 3:
lock.release()
break
t1 = Thread(target=fu, daemon=True)
t2 = Thread(target=fu, daemon=True)
t1.start()
t2.start()
time.sleep(5)
这是您所拥有的简单示例，但是它可以正常工作（除非专门控制哪个线程排在第一位）。
答案 1 :(得分：0)
谢谢。这是代码：
def f(ID, N):
# global lock
if ID == 2:
time.sleep(0.5)
lock.acquire()
lock.acquire()
for i in range(5):
print(f'{N}: {time.ctime(time.time())} \n')
print(f"{i}: lock.locked() = {lock.locked()} \n")
time.sleep(1)
if i == 2 and lock.locked() == True:
lock.release()
有效。在线程1完成i = 2之后，线程2进入for循环。最后，它是lock.locked（）= False。所以这就是我想要的结果，我完全不明白它为什么起作用。
答案 2 :(得分：0)
现在仅使用一个锁即可工作。acquire（）：
def f(ID, N):
# global lock
if ID == 2:
time.sleep(0.5)
lock.acquire()
for i in range(5):
print(f'''{N}: {time.ctime(time.time())}
{i}: lock.locked() = {lock.locked()}
current thread = {threading.current_thread()} \n\n''')
time.sleep(1)
if i == 2 and lock.locked() == True:
lock.release()
print(f'''{N}: {time.ctime(time.time())}
{i}: lock.locked() = {lock.locked()}
current thread = {threading.current_thread()} \n\n''')
然后我得到以下输出：
Thread-1: Thu Sep 27 20:13:52 2018
0: lock.locked() = True
current thread = <myThread2(Thread-1, started 5880)>
main programm: Thu Sep 27 20:13:52 2018
Thread-1: Thu Sep 27 20:13:53 2018
1: lock.locked() = True
current thread = <myThread2(Thread-1, started 5880)>
main programm: Thu Sep 27 20:13:53 2018
Thread-1: Thu Sep 27 20:13:54 2018
2: lock.locked() = True
current thread = <myThread2(Thread-1, started 5880)>
main programm: Thu Sep 27 20:13:54 2018
Thread-2: Thu Sep 27 20:13:55 2018
0: lock.locked() = True
current thread = <myThread2(Thread-2, started 2004)>
Thread-1: Thu Sep 27 20:13:55 2018
2: lock.locked() = False
current thread = <myThread2(Thread-1, started 5880)>
main programm: Thu Sep 27 20:13:55 2018
Thread-1: Thu Sep 27 20:13:55 2018
3: lock.locked() = True
current thread = <myThread2(Thread-1, started 5880)>
Thread-2: Thu Sep 27 20:13:56 2018
1: lock.locked() = True
current thread = <myThread2(Thread-2, started 2004)>
main programm: Thu Sep 27 20:13:56 2018
Thread-1: Thu Sep 27 20:13:56 2018
4: lock.locked() = True
current thread = <myThread2(Thread-1, started 5880)>
Thread-2: Thu Sep 27 20:13:57 2018
2: lock.locked() = True
current thread = <myThread2(Thread-2, started 2004)>
main programm: Thu Sep 27 20:13:57 2018
Thread-1 finished
Thread-2: Thu Sep 27 20:13:58 2018
2: lock.locked() = False
current thread = <myThread2(Thread-2, started 2004)>
main programm: Thu Sep 27 20:13:58 2018
Thread-2: Thu Sep 27 20:13:59 2018
3: lock.locked() = False
current thread = <myThread2(Thread-2, started 2004)>
Thread-2: Thu Sep 27 20:14:00 2018
4: lock.locked() = False
current thread = <myThread2(Thread-2, started 2004)>
Thread-2 finished
8.259018182754517
lock.locked() = False
它提供了所需的输出，但仍然让我感到困惑。现在线程2在线程1释放它之后获得了锁，那么线程1如何继续与线程2同时运行？我仍然不明白为什么我在问题中写的版本没有给出此输出。
答案 3 :(得分：0)
这是代码：
public function view(User $user, Survey $survey)
{
return false
}
这是输出：
t = time.time()
def f1(a,b):
lock2.acquire()
for i in range(5):
print(f'\t {i} \t {a+b} \t THREAD1 \t {time.ctime(time.time())} \n')
time.sleep(1)
lock2.release()
def f2(a,b):
lock2.acquire()
for i in range(5):
print(f'\t {i**2} \t {a+b} \t THREAD2 \t {time.ctime(time.time())} \n')
time.sleep(1)
lock2.release()
class myThread3 (threading.Thread): # create class of threads
def __init__(self, ID, NAME, f, x, y):
super().__init__(target = f, name = NAME, args = (x,y,)) # kein self in super()!!!
# threading.Thread.__init__(self, group=None, target = f, name = NAME, args = (x,y,))
self.threadID = ID
lock2 = threading.Lock()
# Create new threading.Thread objects
t1 = myThread3(101, "Threaddd-1", f1, 5, 5)
t2 = myThread3(102, "Threaddd-2", f2, 1, 1)
# Start new Threads
t1.start()
t2.start()
t1.join()
t2.join()
所以我不明白，为什么t1和t2不能同时运行？？？ t1调用f1而不是f2。 t2调用f2而不是f1。他们不共享变量。所以锁应该没有任何作用，对吧？？？为什么t2被阻塞直到t1释放锁？