Question

double x = 10;
boost::shared_mutex xSharedMutex;


void r() {
    boost::shared_lock<boost::shared_mutex> lock(xSharedMutex);
    for (int i = 0; i < 100; i++) {
        cout << "**** READ **** " << x << endl;
        usleep(200);
    }
}

void w() {
    boost::upgrade_lock<boost::shared_mutex> lock(xSharedMutex);
    for (int i = 0; i < 100; i++) {
        x = i + 12;
        cout << "---- WRITE ---- " << x <<endl;
        usleep(200);
    }

}

int main() {
    boost::thread t1(&r);
    boost::thread t2(&w);

    sleep(3);
}

由于w（）中添加了upgrade_lock，我希望读取和写入顺序完成。但是，读写同时运行。

shared_lock和upgrade_lock的使用是否有误？怎么解决？感谢。

Answer 1

您需要独特的写作所有权：

boost::unique_lock<boost::shared_mutex> lock(xSharedMutex);
for (int i = 0; i < 100; i++) {
    x = i + 12;
    std::cout << "---- WRITE ---- " << x << std::endl;
    usleep(200);
}

或者你可以ad-hoc升级那个锁：

upgrade_lock<shared_mutex> lock(xSharedMutex);
for (int i = 0; i < 100; i++) {
    {
        upgrade_to_unique_lock<shared_mutex> write_lock(lock);
        x = i + 12;
        std::cout << "---- WRITE ---- " << x << std::endl;
    }
    usleep(200);
}

当然，这个程序的输出混合在一起，因为控制台输出不受任何锁定（这是严格的UB）

使用boost :: upgrade_lock和shared_lock实现读/写线程安全性

1 个答案: