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

Question

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

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

而且我最近看到threading.BoundedSemaphore()，它们有什么区别？例如在以下代码片段中（对线程应用限制）：

import threading

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

Answer 1

A Semaphore 可以释放的次数多于获得的次数，这会将其计数器提高到起始值之上。 A BoundedSemaphore can't 高于起始值。

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

线程模块提供了简单的Semaphoreclass。

A Semaphore 提供了一个 non-bounded 计数器，它允许您调用 release() 任意次数来递增。

但是，为了避免编程错误，使用 BoundedSemaphore 通常是正确的选择，如果 release() 调用试图将计数器增加到超过其最大大小，则会引发错误。

编辑

信号量有一个内部计数器而不是锁定标志（在锁的情况下），并且它只会在超过给定数量的线程试图持有信号量时阻塞。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。

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

What is the difference between .Semaphore() and .BoundedSemaphore()?

python

multithreading

mutex

semaphore

python-multithreading