C++11多线程编程-两个进程轮流打印1~100

这是经典的同步互斥问题，

遵循原则：

1、条件变量需要锁的保护；

2、锁需要条件变量成立后，后重新上锁；

参考代码：

//notify_one()（随机唤醒一个等待的线程）//notify_all()（唤醒所有等待的线程）//Create By@herongwei /09/10#include <bits/stdc++.h>#include <mutex>#include <thread>#include <condition_variable>using namespace std;std::mutex data_mutex;//互斥锁std::condition_variable data_var;//条件变量bool flag = true;void printfA() {int i = 1;while(i <= 100) {//休息1秒//std::this_thread::sleep_for(std::chrono::seconds(1));std::unique_lock<std::mutex> lck(data_mutex);data_var.wait(lck,[]{return flag;});//等待flag=true才打印奇数std::cout<<"A " << i <<endl;i += 2;flag = false;data_var.notify_one();}}void printfB() {int i = 2;while(i <= 100) {std::unique_lock<std::mutex> lck(data_mutex);data_var.wait(lck,[]{return !flag;});//等待flag=false才打印偶数std::cout<<"B " << i <<endl;i += 2;flag = true;data_var.notify_one();}}int main() {// freopen("in.txt","r",stdin);std::thread tA(printfA);std::thread tB(printfB);tA.join();tB.join();return 0;}

思路2:

C++ 11，VS

使用了unique_lock，条件变量

条件变量的wait(), notify_one()

wait()中使用了lambda表达式

代码

#include <iostream>

#include <vector>

#include <thread>

#include <list>

#include <mutex>

using namespace std;

bool flag = true;

class A

{

public:

void print1()

{

while (true)

{

std::unique_lock<std::mutex> lock1(print_mutex);

data_var.wait(lock1, [] {return flag; });

cout << "thread: " << std::this_thread::get_id() << " print: " << "A" << endl;

flag = false;

data_var.notify_one();

}

}

void print2()

{

while (true)

{

std::this_thread::sleep_for(std::chrono::seconds(1));

std::unique_lock<std::mutex> lock2(print_mutex);

data_var.wait(lock2, [] {return !flag; });

cout << "thread: " << std::this_thread::get_id() << " print: " << "B" << endl;

flag = true;

data_var.notify_one();

}

}

private:

std::mutex print_mutex;

std::condition_variable data_var;

};

int main()

{

A myobj_a;

std::thread print1(&A::print1, &myobj_a);

std::thread print2(&A::print2, &myobj_a);

print1.join();

print2.join();

return 0;

}