c++ - C++ 11多线程生产者/消费者程序挂起

Question

我是C++ 11的新手，并使用线程功能。在以下程序中，主线程启动9个工作线程，并将数据插入队列，然后等待线程终止。我看到工作线程没有被唤醒，程序只是挂起了。

#include <iostream>
#include <thread>
#include <mutex>
#include <condition_variable>
#include <queue>
#include <vector>
#include <chrono>
#include <future>
#include <atomic>
using namespace std::chrono_literals;

std::mutex _rmtx;
std::mutex _wmtx;
std::queue<unsigned long long> dataq;
std::condition_variable _rcv;
std::condition_variable _wcv;
std::atomic_bool termthd;

void thdfunc(const int& num)
{
    std::cout << "starting thread#" << num << std::endl;

    std::unique_lock<std::mutex> rul(_rmtx);
    while (true) {
        while(!_rcv.wait_until(rul, std::chrono::steady_clock::now() + 10ms, [] {return !dataq.empty() || termthd.load(); }));
        if (termthd.load()) {
            std::terminate();
        }
        std::cout<<"thd#" << num << " : " << dataq.front() <<std::endl;
        dataq.pop();
        _wcv.notify_one();
    }
}

int main()
{
    std::vector<std::thread*> thdvec;
    std::unique_lock<std::mutex> wul(_rmtx);
    unsigned long long data = 0ULL;
    termthd.store(false);
    for (int i = 0; i < 9; i++) {
        thdvec.push_back(new std::thread(thdfunc, i));
    }

    for ( data = 0ULL; data < 2ULL; data++) {
        _wcv.wait_until(wul, std::chrono::steady_clock::now() + 10ms, [&] {return data > 1000000ULL; });
        dataq.push(std::ref(data));
        _rcv.notify_one();
    }
    termthd.store(true);
    _rcv.notify_all();
    //std::this_thread::yield();
    for (int i = 0; i < 9; i++) {
        thdvec[i]->join();
    }
}

我无法找出问题所在。如何确保线程被唤醒并处理请求并正常终止？

Answer 1

此std::unique_lock<std::mutex> wul(_rmtx);将锁定_rmtx互斥锁，直到主作用域结束。这肯定是一个问题，因为其他试图锁定_rmtx的线程将阻塞:

int main()
{
    std::vector<std::thread*> thdvec;
    std::unique_lock<std::mutex> wul(_rmtx); // <- locking mutex until end of main. 
    // other threads trying to lock _rmtx will block

    unsigned long long data = 0ULL;
    // ... rest of the code ...