What Is The Difference Between .semaphore() And .boundedsemaphore()?
Solution 1:
A Semaphore
can be released more times than it's acquired, and that will raise its counter above the starting value. A BoundedSemaphore
can't be raised above the starting value.
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.')
Solution 2:
The threading module provides the simple Semaphore
class.
A Semaphore
provides a non-bounded counter which allows you to call release()
any number of times for incrementing.
However, to avoid programming errors, it’s usually a correct choice to use BoundedSemaphore
, which raises an error if a release()
call tries to increase the counter beyond its maximum size.
EDIT
A semaphore has an internal counter rather than a lock flag (in case of Locks), and it only blocks if more than a given number of threads have attempted to hold the semaphore. Depending on how the semaphore is initialized, this allows multiple threads to access the same code section simultaneously.
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