.Semaphore（）和.BoundedSemaphore（）有什么区别？

Question

我知道threading.Lock()等于threading.Semaphore(1)。在

threading.Lock()是否也等于threading.BoundedSemaphore(1)？在

我刚认识threading.BoundedSemaphore()，这两个有什么区别？例如以下代码片段（对线程应用限制）：

import threading

sem = threading.Semaphore(5)
sem = threading.BoundedSemaphore(5)

Answer 1

一个Semaphore的释放次数可能比获得的次数多，这将使其计数器高于起始值。一个BoundedSemaphorecan't高于起始值。在

from threading import Semaphore, BoundedSemaphore

# Usually, you create a Semaphore that will allow a certain number of threads
# into a section of code. This one starts at 5.
s1 = Semaphore(5)

# When you want to enter the section of code, you acquire it first.
# That lowers it to 4. (Four more threads could enter this section.)
s1.acquire()

# Then you do whatever sensitive thing needed to be restricted to five threads.

# When you're finished, you release the semaphore, and it goes back to 5.
s1.release()


# That's all fine, but you can also release it without acquiring it first.
s1.release()

# The counter is now 6! That might make sense in some situations, but not in most.
print(s1._value)  # => 6

# If that doesn't make sense in your situation, use a BoundedSemaphore.

s2 = BoundedSemaphore(5)  # Start at 5.

s2.acquire()  # Lower to 4.

s2.release()  # Go back to 5.

try:
    s2.release()  # Try to raise to 6, above starting value.
except ValueError:
    print('As expected, it complained.')    

Answer 2

线程模块提供简单的Semaphore类。在

Semaphore提供了一个无界计数器，允许您调用release()任意次数进行递增。在

但是，为了避免编程错误，使用BoundedSemaphore通常是一个正确的选择，如果release()调用试图将计数器增加到其最大大小之外，则会引发错误。在

编辑

一个信号量有一个内部计数器而不是一个锁标志（在锁的情况下），它只有在超过给定数量的线程试图保持信号量时才会阻塞。根据信号量的初始化方式，这允许多个线程同时访问同一代码段。在