python学习第37天GIL锁、死锁现象与递归锁、信号量、Event时间、线程queue

定义：
In CPython, the global interpreter lock, or GIL, is a mutex that prevents multiple 
native threads from executing Python bytecodes at once. This lock is necessary mainly 
because CPython’s memory management is not thread-safe. (However, since the GIL 
exists, other features have grown to depend on the guarantees that it enforces.)
‘‘‘
结论：在Cpython解释器中，同一个进程下开启的多线程，同一时刻只能有一个线程执行，无法利用多核优势

#1. cpu到底是用来做计算的，还是用来做I/O的？

#2. 多cpu，意味着可以有多个核并行完成计算，所以多核提升的是计算性能

#3. 每个cpu一旦遇到I/O阻塞，仍然需要等待，所以多核对I/O操作没什么用处 

分析：
我们有四个任务需要处理，处理方式肯定是要玩出并发的效果，解决方案可以是：
方案一：开启四个进程
方案二：一个进程下，开启四个线程

#单核情况下，分析结果: 
　　如果四个任务是计算密集型，没有多核来并行计算，方案一徒增了创建进程的开销，方案二胜
　　如果四个任务是I/O密集型，方案一创建进程的开销大，且进程的切换速度远不如线程，方案二胜

#多核情况下，分析结果：
　　如果四个任务是计算密集型，多核意味着并行计算，在python中一个进程中同一时刻只有一个线程执行用不上多核，方案一胜
　　如果四个任务是I/O密集型，再多的核也解决不了I/O问题，方案二胜

 
#结论：现在的计算机基本上都是多核，python对于计算密集型的任务开多线程的效率并不能带来多大性能上的提升，甚至不如串行(没有大量切换)，但是，对于IO密集型的任务效率还是有显著提升的。
复制代码

计算密集型:应该使用多进程
# from multiprocessing import Process
# from threading import Thread
# import os,time
#
# def work():
#     res=0
#     for i in range(100000000):
#         res*=i
#
# if __name__ == ‘__main__‘:
#     l=[]
#     print(os.cpu_count())
#     start=time.time()
#     for i in range(6):
#         # p=Process(target=work)
#         p=Thread(target=work)
#         l.append(p)
#         p.start()
#     for p in l:
#         p.join()
#     stop=time.time()
#     print(‘run time is %s‘ %(stop-start)) #4.271663427352905

# IO密集型: 应该开启多线程
from multiprocessing import Process
from threading import Thread
import threading
import os,time
def work():
    time.sleep(2)

if __name__ == ‘__main__‘:
    l=[]
    start=time.time()
    for i in range(300):
        # p=Process(target=work) #2.225289821624756
        p=Thread(target=work) #2.002105951309204
        l.append(p)
        p.start()
    for p in l:
        p.join()
    stop=time.time()
    print(‘run time is %s‘ %(stop-start))

from threading import Thread,Lock
import time

mutex=Lock()
n=100
def task():
    global n
    with mutex:
        temp=n
        time.sleep(0.1)
        n=temp-1

if __name__ == ‘__main__‘:
    l=[]
    for i in range(100):
        t=Thread(target=task)
        l.append(t)
        t.start()

    for t in l:
        t.join()
    print(n)#结果可能是99，代码运行顺序错乱

from threading import Thread,Lock
import os,time
def work():
    global n
    lock.acquire()
    temp=n
    time.sleep(0.1)
    n=temp-1
    lock.release()
if __name__ == ‘__main__‘:
    lock=Lock()
    n=100
    l=[]
    for i in range(100):
        p=Thread(target=work)
        l.append(p)
        p.start()
    for p in l:
        p.join()

    print(n) #结果肯定为0，由原来的并发执行变成串行，牺牲了执行效率保证了数据安全
复制代码

#不加锁:并发执行,速度快,数据不安全
from threading import current_thread,Thread,Lock
import os,time
def task():
    global n
    print(‘%s is running‘ %current_thread().getName())
    temp=n
    time.sleep(0.5)
    n=temp-1


if __name__ == ‘__main__‘:
    n=100
    lock=Lock()
    threads=[]
    start_time=time.time()
    for i in range(100):
        t=Thread(target=task)
        threads.append(t)
        t.start()
    for t in threads:
        t.join()

    stop_time=time.time()
    print(‘主:%s n:%s‘ %(stop_time-start_time,n))

‘‘‘
Thread-1 is running
Thread-2 is running
......
Thread-100 is running
主:0.5216062068939209 n:99
‘‘‘

#不加锁:未加锁部分并发执行,加锁部分串行执行,速度慢,数据安全
from threading import current_thread,Thread,Lock
import os,time
def task():
    #未加锁的代码并发运行
    time.sleep(3)
    print(‘%s start to run‘ %current_thread().getName())
    global n
    #加锁的代码串行运行
    lock.acquire()
    temp=n
    time.sleep(0.5)
    n=temp-1
    lock.release()

if __name__ == ‘__main__‘:
    n=100
    lock=Lock()
    threads=[]
    start_time=time.time()
    for i in range(100):
        t=Thread(target=task)
        threads.append(t)
        t.start()
    for t in threads:
        t.join()
    stop_time=time.time()
    print(‘主:%s n:%s‘ %(stop_time-start_time,n))

‘‘‘
Thread-1 is running
Thread-2 is running
......
Thread-100 is running
主:53.294203758239746 n:0
‘‘‘

#有的同学可能有疑问:既然加锁会让运行变成串行,那么我在start之后立即使用join,就不用加锁了啊,也是串行的效果啊
#没错:在start之后立刻使用jion,肯定会将100个任务的执行变成串行,毫无疑问,最终n的结果也肯定是0,是安全的,但问题是
#start后立即join:任务内的所有代码都是串行执行的,而加锁,只是加锁的部分即修改共享数据的部分是串行的
#单从保证数据安全方面,二者都可以实现,但很明显是加锁的效率更高.
from threading import current_thread,Thread,Lock
import os,time
def task():
    time.sleep(3)
    print(‘%s start to run‘ %current_thread().getName())
    global n
    temp=n
    time.sleep(0.5)
    n=temp-1


if __name__ == ‘__main__‘:
    n=100
    lock=Lock()
    start_time=time.time()
    for i in range(100):
        t=Thread(target=task)
        t.start()
        t.join()
    stop_time=time.time()
    print(‘主:%s n:%s‘ %(stop_time-start_time,n))

‘‘‘
Thread-1 start to run
Thread-2 start to run
......
Thread-100 start to run
主:350.6937336921692 n:0 #耗时是多么的恐怖
‘‘‘

复制代码
from threading import Thread,Lock
import time
mutexA=Lock()
mutexB=Lock()

class MyThread(Thread):
    def run(self):
        self.func1()
        self.func2()
    def func1(self):
        mutexA.acquire()
        print(‘\033[41m%s 拿到A锁\033[0m‘ %self.name)

        mutexB.acquire()
        print(‘\033[42m%s 拿到B锁\033[0m‘ %self.name)
        mutexB.release()

        mutexA.release()

    def func2(self):
        mutexB.acquire()
        print(‘\033[43m%s 拿到B锁\033[0m‘ %self.name)
        time.sleep(2)

        mutexA.acquire()
        print(‘\033[44m%s 拿到A锁\033[0m‘ %self.name)
        mutexA.release()

        mutexB.release()

if __name__ == ‘__main__‘:
    for i in range(10):
        t=MyThread()
        t.start()

‘‘‘
Thread-1 拿到A锁
Thread-1 拿到B锁
Thread-1 拿到B锁
Thread-2 拿到A锁
然后就卡住，死锁了

from threading import Thread,Lock,RLock
import time

# mutexA=Lock()
# mutexB=Lock()
mutexB=mutexA=RLock()


class Mythead(Thread):
    def run(self):
        self.f1()
        self.f2()

    def f1(self):
        mutexA.acquire()
        print(‘%s 抢到A锁‘ %self.name)
        mutexB.acquire()
        print(‘%s 抢到B锁‘ %self.name)
        mutexB.release()
        mutexA.release()

    def f2(self):
        mutexB.acquire()
        print(‘%s 抢到了B锁‘ %self.name)
        time.sleep(2)
        mutexA.acquire()
        print(‘%s 抢到了A锁‘ %self.name)
        mutexA.release()
        mutexB.release()

if __name__ == ‘__main__‘:
    for i in range(100):
        t=Mythead()
        t.start()

from threading import Thread,Semaphore
import threading
import time
# def func():
#     if sm.acquire():
#         print (threading.currentThread().getName() + ‘ get semaphore‘)
#         time.sleep(2)
#         sm.release()
def func():
    sm.acquire()
    print(‘%s get sm‘ %threading.current_thread().getName())
    time.sleep(3)
    sm.release()
if __name__ == ‘__main__‘:
    sm=Semaphore(5)
    for i in range(23):
        t=Thread(target=func)
        t.start()

from threading import Thread,Event
import threading
import time,random
def conn_mysql():
    count=1
    while not event.is_set():
        if count > 3:
            raise TimeoutError(‘链接超时‘)
        print(‘<%s>第%s次尝试链接‘ % (threading.current_thread().getName(), count))
        event.wait(0.5)
        count+=1
    print(‘<%s>链接成功‘ %threading.current_thread().getName())


def check_mysql():
    print(‘\033[45m[%s]正在检查mysql\033[0m‘ % threading.current_thread().getName())
    time.sleep(random.randint(2,4))
    event.set()
if __name__ == ‘__main__‘:
    event=Event()
    conn1=Thread(target=conn_mysql)
    conn2=Thread(target=conn_mysql)
    check=Thread(target=check_mysql)

    conn1.start()
    conn2.start()
    check.start()
复制代码

import queue

# queue.Queue() #先进先出
# q=queue.Queue(3)
# q.put(1)
# q.put(2)
# q.put(3)
# print(q.get())
# print(q.get())
# print(q.get())

# queue.LifoQueue() #后进先出->堆栈
# q=queue.LifoQueue(3)
# q.put(1)
# q.put(2)
# q.put(3)
# print(q.get())
# print(q.get())
# print(q.get())
# queue.PriorityQueue() #优先级
q=queue.PriorityQueue(3) #优先级,优先级用数字表示,数字越小优先级越高
q.put((10,‘a‘))
q.put((-1,‘b‘))
q.put((100,‘c‘))
print(q.get())
print(q.get())
print(q.get())