通过N个线程顺序循环打印从0至100

为什么80%的码农都做不了架构师？>>>   

如：

通过N个线程顺序循环打印从0至100，如给定N=3则输出:
thread0: 0
thread1: 1
thread2: 2
thread0: 3
thread1: 4
.....

解：N个线程对应N个条件变量，依次激活下一个线程

#include 
#include 
#include 
#include 
#include 

constexpr int THREADS = 3;
constexpr int MAX = 100;

std::mutex m;
std::condition_variable cv[THREADS];
bool flags[THREADS];

int n = 0;

void f(int i)
{
    const int next = (i + 1) % THREADS;
    for (;;)
    {
        std::unique_lock lock(m);
        cv[i].wait(lock, [=](){return flags[i];});
        if (n <= MAX)
            std::cout << "thread " << i << ": " << n++ << std::endl;
        flags[i] = false;
        flags[next] = true;
        lock.unlock();
        cv[next].notify_one();
        if (n > MAX)
            break;
    }
}

int main()
{
    flags[0] = true;
    std::vector vec;
    for (int i = 0; i < THREADS; ++i)
        vec.emplace_back(f, i);
    for (auto &t: vec)
        t.join();
    return 0;
}

 

转载于:https://my.oschina.net/guzhou/blog/3023693