semaphore信號量的簡單代碼演示

import threading
import logging
import time

FORMAT = '%(threadName)s %(thread)d %(message)s'
logging.basicConfig(format=FORMAT, level=logging.INFO)

def worker(s:threading.Semaphore):
    logging.info('in sub thread')
    logging.info(s.acquire())   # 獲取信號量，計數(shù)器 -1
    logging.info('sub thread over')

s = threading.Semaphore(3)   # 創(chuàng)建3個信號量計數(shù)器
logging.info(s.acquire())
print(s._value)    # 看看現(xiàn)在信號量的內(nèi)的數(shù)值是多少
logging.info(s.acquire())
print(s._value)
logging.info(s.acquire())
print(s._value)

threading.Thread(target=worker, args=(s, )).start()

time.sleep(2)

logging.info(s.acquire(False))    # 不阻塞，若獲取不到信號量，則為False
logging.info(s.acquire(timeout=10))   # 設(shè)置超時時間，等待超時時間過了還未獲取到信號，返回值為False

# 釋放
logging.info('released')
s.release()   # 釋放信號量，計數(shù)器i + 1

簡單的資源池演示

import threading
import logging
import random

FORMAT = '%(threadName)s %(thread)d %(message)s'
logging.basicConfig(format=FORMAT, level=logging.INFO)

class Conn:
    def __init__(self, name):
        self.name = name

    def __str__(self):
        return self.name

class Pool:
    def __init__(self, count:int):
        self.count = count
        self.pool = [ self._connect('conn-{}'.format(x)) for x in range(self.count)]

    def _connect(self, conn_name):
        return Conn(conn_name)

    def get_conn(self):
        conn = self.pool.pop()
        return conn

    def return_conn(self, conn:Conn):
        self.pool.append(conn)

pool = Pool(3)

def worker(pool:Pool):
    conn = pool.get_conn()
    logging.info(conn)
    threading.Event().wait(random.randint(1,4))
    pool.return_conn(conn)

for i in range(6):
    threading.Thread(target=worker, name='worker-{}'.format(i), args=(pool,)).start()       

使用semaphore來完善代碼

import threading
import logging
import random

FORMAT = '%(threadName)s %(thread)d %(message)s'
logging.basicConfig(format=FORMAT, level=logging.INFO)

class Conn:
    def __init__(self, name):
        self.name = name

    def __str__(self):
        return self.name

class Pool:
    def __init__(self, count:int):
        self.count = count
        self.pool = [ self._connect('conn-{}'.format(x)) for x in range(self.count)]
        self.semahore = threading.Semaphore(count)

    def _connect(self, conn_name):
        return Conn(conn_name)

    def get_conn(self):
        self.semahore.acquire()
        conn = self.pool.pop()
        return conn

    def return_conn(self, conn:Conn):
        self.pool.append(conn)
        self.semahore.release()

pool = Pool(3)

def worker(pool:Pool):
    conn = pool.get_conn()
    logging.info(conn)
    threading.Event().wait(random.randint(1,4))
    pool.return_conn(conn)

for i in range(6):
    threading.Thread(target=worker, name='worker-{}'.format(i), args=(pool,)).start()

關(guān)于信號量release超出初始值范圍的問題

當我們使用semaphore的時候，如果還未acquire，就release了，會產(chǎn)生什么問題？
會產(chǎn)生是的信號量的值+1，超出了原本設(shè)置semaphore的初始值，下面的例子說明了這個問題

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import threading
import logging

FORMAT = '%(threadName)s %(thread)d %(message)s'
logging.basicConfig(format=FORMAT, level=logging.INFO)

sema = threading.Semaphore(3)
logging.warning(sema.__dict__)
for _ in range(3):
    sema.acquire()

logging.warning('-----')
logging.warning(sema.__dict__)

for _ in range(4):
    sema.release()
logging.warning(sema.__dict__)

for _ in range(3):
    sema.acquire()
logging.warning('--------')
logging.warning(sema.__dict__)
sema.acquire()
logging.warning('======')
logging.warning(sema.__dict__)

所以這個這樣的問題，可以使用BoundedSemaphore類來實現(xiàn)有界的信號量，若relase超出了初始值的范圍，會拋出ValueError異常