各种线程的unique_lock

Question

我有一个消费者和两个生产者。当我同时生成两个生产者时，他们似乎互相锁定，因为我们看到的第一个值是223和889。

有人可以解释一下这里发生了什么吗？

#include<vector>
#include<thread>
#include<iostream>
#include<mutex>
#include<chrono>
#include <condition_variable>

using namespace std;

vector<double>testvec;
mutex mtx;
condition_variable cv;

class Base
{
public:
    Base() {};
    void dosomething();
    int i;
};

void Base::dosomething()
{
    while(1)
    {
        std::unique_lock<std::mutex> ulck(mtx);
        testvec.push_back(i);
        ulck.unlock();
        cv.notify_all();
        i++;
        std::this_thread::sleep_for (std::chrono::milliseconds(i));
    }
};

class Derived1 : public Base
{
public:
    Derived1() {i = 222;}
};

class Derived2 : public Base
{
public:
    Derived2() {i = 888;}
};

class Consumer
{
public:
    Consumer() {}
    void dostuff();
};

void Consumer::dostuff()
{
    while(1)
    {
        std::unique_lock<std::mutex> ulck(mtx);    //locks shared data
        cv.wait(ulck);
        cout<<"last value: "<<testvec.back()<<endl;
        ulck.unlock();
    }
}

int main( int argc, char ** argv )
{
    Derived1 derived1;
    Derived2 derived2;
    Consumer c;

    std::thread t1(&Derived1::dosomething, &derived1);
    std::thread t2(&Derived2::dosomething, &derived2);
    std::thread t3(&Consumer::dostuff, &c);

    t1.join();
    t2.join();
    t3.join();
}

output is:
last value: 223
last value: 224
last value: 225
last value: 889
last value: 226
last value: 227
last value: 228
last value: 229
last value: 890
last value: 230

expected output:
last value: 888 (or 222)
last value: 222 (or 888)
last value: 223
...

Answer 1

没有谓词，你永远不会从wait得到理智的行为。另外，为什么你继续释放锁定只是为了立即再锁定它？

这是你需要做的：

void Consumer::dostuff()
{
    std::unique_lock<std::mutex> ulck(mtx);    //locks shared data
    int i = 0;
    while(1)
    {

        // Correctly decide when to wait and when to stop waiting
        while (testvec.empty() || (testvec.back() == i))
            cv.wait(ulck);

        i = testvec.back();
        cout<<"last value: "<<testvec.back()<<endl;
    }
}

条件变量是无状态的，无法知道它们是否应该等待或何时应该停止等待，因为它们是共享状态的函数。使用互斥锁来保护共享状态是您的责任。

请注意，如果必须等待，此代码仅调用wait。您只能通过检查共享状态来判断是否需要等待。并注意它继续调用wait，而它必须等待。同样，您只能通过检查共享状态来判断是否可以停止等待。共享状态是你的责任 - 条件变量是无状态的。

Answer 2

我有一个消费者和两个生产者。

错误。你有2个生产者和一个采样器。什么都没有消耗。

你可能想要的是这个（简化之后）：

#include<deque>
#include<thread>
#include<iostream>
#include<mutex>
#include<chrono>
#include <condition_variable>

using namespace std;

deque<double> testvec;
mutex mtx;
condition_variable cv;

auto producer = [](int i)
{
    while(1)
    {
        std::unique_lock<std::mutex> ulck(mtx);
        testvec.push_back(i);
        ulck.unlock();
        cv.notify_all();
        i++;
        std::this_thread::sleep_for (std::chrono::milliseconds(i));
    }
};

void consume()
{
    while(1)
    {
        std::unique_lock<std::mutex> ulck(mtx);    //locks shared data


        // wait until there is something to consume...
        cv.wait(ulck, [&]{return not testvec.empty(); });

        /// ... consume it...
        auto val = testvec.front();
        testvec.pop_front();

        /// ... now unlock and work with the consumed data while unlocked
        ulck.unlock();

        /// ... do work
        cout<<"last value: " << val <<endl;
    }
}

int main( int argc, char ** argv )
{
    std::thread t1(producer, 222);
    std::thread t2(producer, 888);
    std::thread t3(consume);

    t1.join();
    t2.join();
    t3.join();
}

示例输出：

last value: 222
last value: 888
last value: 223
last value: 224
last value: 225
last value: 889
last value: 226
...