threading 之 semaphore信号量

Posted 2021-02-10

semaphore信号量的简单代码演示

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())   # 获取信号量，计数器 -1
    logging.info(‘sub thread over‘)

s = threading.Semaphore(3)   # 创建3个信号量计数器
logging.info(s.acquire())
print(s._value)    # 看看现在信号量的内的数值是多少
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，若获取不到信号量，则为False
logging.info(s.acquire(timeout=10))   # 设置超时时间，等待超时时间过了还未获取到信号，返回值为False

# 释放
logging.info(‘released‘)
s.release()   # 释放信号量，计数器i + 1

简单的资源池演示

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来完善代码

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()

关于信号量release超出初始值范围的问题

当我们使用semaphore的时候，如果还未acquire，就release了，会产生什么问题？
会产生是的信号量的值+1，超出了原本设置semaphore的初始值，下面的例子说明了这个问题

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__)

所以这个这样的问题，可以使用BoundedSemaphore类来实现有界的信号量，若relase超出了初始值的范围，会抛出ValueError异常