02-02-01

转载

mob604756ec5243 2021-07-13 09:51:00

不管进程还是线程，主都会等着子结束而结束这是因为子进程/线程在执行的时候，会暂用一些资源，主就要收回这些资源

线程是系统调度的基本单位，进程是系统资源的基本单位

主线程会等待子线程结束才结束

import threading
import time

def sayHello():
    print("------")
    time.sleep(1)


for i in range(5):
    t = threading.Thread(target=sayHello)
    t.start()

print("*******")
Result：

------
------
------
------
------

******* #多个线程执行同一个函数，不会互相影响

#####################使用thread子类完成创建多线程###########
import threading
import time

class myThread(threading.Thread):
    def run(self):
        for i in range(3):
            time.sleep(1)
            print(self.name+"num"+str(i)) #self.name保存的是当前的线程名字


if __name__ == "__main__":
    t = myThread()
    t.start() #MyThread没有start方法，为啥可以使用呢，这是threading.Thread中start去调用myThread的run方法

线程的执行顺序：

线程的执行顺序没有固定的，各个线程执行顺序不固定

线程共享全局变量：

线程全局变量是共用的，一个线程的更改其它线程使用也是更改的；而进程是相互独立的，一个进程修改不会对另一个进程产生影响

import time
import threading
g_num = 0

def work1():
    global  g_num
    g_num += 1
    print("work1, gum is %d"%g_num)

def work2():
    global  g_num
    print("work2, gum is %d"%g_num)

t1 = threading.Thread(target = work1)
t1.start()
time.sleep(1)
t2 = threading.Thread(target = work2)
t2.start()
result:

work1, gum is 1
work2, gum is 1

列表作为参数传递也可以共享，变量不行需要global
import time
import threading

def work1(g_num):
    g_num.append(4)
    print("work1, guum is", g_num)

def work2(g_num):
    time.sleep(1)
    print("work2, guum is", g_num)

g_num = [1,2,3]
t1 = threading.Thread(target = work1, args=(g_num,))
t1.start()
t2 = threading.Thread(target = work2, args=(g_num,))
t2.start()

Result:

work1, guum is [1, 2, 3, 4]
work2, guum is [1, 2, 3, 4]

####互斥锁

当多个线程同时竞争同一个资源的的时候

互斥锁是加的code范围越小越好，加的太多越来越像单任务

比如下面如果在for循环外面加就跟单任务很相似

import threading

g_num = 0

def work1():
    global g_num
    #这个线程和work2都在抢着对这个锁进行上锁，如果一方的锁上锁成功，那么另一方需要需要一直等待，知道这个锁释放为
    for i in range(1000000):
        mutex.acquire()
        g_num += 1
        mutex.release() #解锁
    print("work1, guum is %d"%g_num)

def work2():
    global g_num
    for i in range(1000000):
        mutex.acquire()
        g_num += 1
        mutex.release()
    print("work1, guum is %d"%g_num)

#创建一把锁，这个锁默认是没有上锁的
mutex = threading.Lock()
t1 = threading.Thread(target = work1)
t1.start()
#t1.join()
t2 = threading.Thread(target = work2)
t2.start()
#t2.join()

多线程使用非共享变量

创建一个函数，多线程访问一个函数时，每个线程访问这个函数的变量值时独有的

import threading
import time


def work1():

    name = threading.current_thread().name #取线程名字比如 Thread-1 Main thread...
    print("---thread name---%s"%name)
    num = 0
    if name == "Thread-1":
        num += 1
    else:
        time.sleep(1)
    print("thread name is %s---num is %d"%(name, num))


t1 = threading.Thread(target = work1)
t1.start()
#t1.join()
t2 = threading.Thread(target = work1)
t2.start()
#t2.join()
result:

---thread name---Thread-1
thread name is Thread-1---num is 1
---thread name---Thread-2
thread name is Thread-2---num is 0

死锁以及解决方法

加入超时时间：

 mutex.acquire(2)###2s之后还不能上锁的化，就继续往下走

同步的理解与应用
代码按照一定规律去执行, 不然很难保证task1 task2 task3依次输出

import threading
import time

class Task1(threading.Thread):
    def run(self):
        while True:
            if lock1.acquire():
                print("----Task1----")
                time.sleep(1)
                lock2.release()

class Task2(threading.Thread):
    def run(self):
        while True:
            if lock2.acquire():
                print("----Task2----")
                time.sleep(1)
                lock3.release()

class Task3(threading.Thread):
    def run(self):
        while True:
            if lock3.acquire():
                print("----Task3----")
                time.sleep(1)
                lock1.release()


if __name__ == "__main__":
    lock1 = threading.Lock()
    lock2 = threading.Lock()
    lock2.acquire()
    lock3 = threading.Lock()
    lock3.acquire()
    t1 = Task1()
    t2 = Task2()
    t3 = Task3()
    t1.start()
    t2.start()
    t3.start()
result: