.Semaphore()和.BoundedSemaphore()有什么区别？
.Semaphore()和.BoundedSemaphore()有什么区别？
提问于 2018-02-25 15:08:18
我知道threading.Lock()等同于threading.Semaphore(1)。
threading.Lock()是否也等同于threading.BoundedSemaphore(1)？
最近我遇到了threading.BoundedSemaphore()，它们之间有什么不同？例如下面的代码片段(对线程应用限制)：
import threading
sem = threading.Semaphore(5)
sem = threading.BoundedSemaphore(5)回答 1
Stack Overflow用户
回答已采纳
发布于 2018-02-25 15:12:39
一个Semaphore被释放的次数可以比它获得的次数多，这会使它的计数器超过起始值。BoundedSemaphore can't将提升到起始值以上。
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.') 页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/48971121
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