Question

我正在尝试使用线程池更新向量，其中每个工作者负责更新向量的某些部分。更具体地说，我将矢量分割成具有空交集的子集，其中集合的数量由我的计算机可用的cpus数量给出（4）。

每个线程工作者都会更新：

std::array<int, arraySize> vec

如果在类MultiThreadedUpdateOfVector中将此数组声明为私有，则该功能不起作用：每次交易都会更新vec的相应部分，但下一个线程不会选择更改。因此，vec就好像是每个线程的局部变量。

如果出现以下情况，此问题就会消失：

std::array<int, arraySize> vec

在MultiThreadedUpdateOfVector之前声明。

你能解释一下这种不受欢迎的行为吗？

您能否建议std::array<int, arraySize> vec仍然是MultiThreadedUpdateOfVector成员的解决方案？

谢谢！

#include "stdafx.h"
#include <iostream>
#include <vector>
#include <chrono>
#include <thread>
#include <array>

#include "ThreadPool.h"
using namespace std;

const int block = 2;
const int arraySize = 8;

class MultiThreadedUpdateOfVector
{
public:
    MultiThreadedUpdateOfVector()
    {

    }
    bool setArray(const int& i, const int& j)
    {
        std::thread::id id = std::this_thread::get_id();
        for (int kk = i; kk < j; ++kk)
        {
            vec[kk] = i * 10000 + j * 100 + kk;
        }
        return true;
    }
    void print()
    {
        for (unsigned kk = 0; kk < vec.size(); ++kk)
        {
            std::cout << kk << "     " << vec[kk] << endl;
        }
    }
private:
    std::array<int, arraySize> vec;
};

int main()
{
    std::thread::id id = std::this_thread::get_id();
    ThreadPool pool(4);
    std::vector<std::future<bool> >results;

    MultiThreadedUpdateOfVector h;
    int begin = 0;
    int end = block;
    for (int i = 0; i < 4; ++i) {
           results.push_back(pool.enqueue(&MultiThreadedUpdateOfVector::setArray, h, begin, end));
        begin = end + 1;
        end += block;
    }
    for (int i = 0; i < 4; ++i)
        results[i].get();
    h.print();
    return 0;
}

#include <vector>
#include <queue>
#include <memory>
#include <thread>
#include <mutex>
#include <condition_variable>
#include <future>
#include <functional>
#include <stdexcept>

class ThreadPool {
public:
    ThreadPool(size_t);
    template<class F, class... Args>
    auto enqueue(F&& f, Args&&... args)
        ->std::future<typename std::result_of<F(Args...)>::type>;
    ~ThreadPool();
    int getTasksSize()
    {
        std::unique_lock<std::mutex> lock(this->queue_mutex_m);
        std::thread::id id1 = std::this_thread::get_id();
        return tasks_m.size();
    }
private:
    // need to keep track of threads so we can join them
    std::vector< std::thread > workers_m;
    // the task queue
    std::queue< std::function<void()> > tasks_m;

    // synchronization
    std::mutex queue_mutex_m;
    std::condition_variable condition_m;
    bool stop_m;
};

// the constructor just launches some amount of workers
inline ThreadPool::ThreadPool(size_t threads)
    : stop_m(false)
{
    std::thread::id id = std::this_thread::get_id();
    for (size_t i = 0; i < threads; ++i)
    {
        workers_m.emplace_back(
            [this]
        {
            for (;;)
            {
                std::function<void()> task;
                {
                    std::unique_lock<std::mutex> lock(this->queue_mutex_m);
                    std::thread::id id1 = std::this_thread::get_id();
                    this->condition_m.wait(lock, [this]{ return this->stop_m || !this->tasks_m.empty(); });
                    std::thread::id id = std::this_thread::get_id();
                    if (this->stop_m && this->tasks_m.empty())
                        return;
                    task = std::move(this->tasks_m.front());
                    this->tasks_m.pop();
                }
                task();
            }
        }
        );
    }
}

// add new work item to the pool
template<class F, class... Args>
auto ThreadPool::enqueue(F&& f, Args&&... args)
-> std::future<typename std::result_of<F(Args...)>::type>
{
    std::thread::id id = std::this_thread::get_id();
    using return_type = typename std::result_of<F(Args...)>::type;

    auto task = std::make_shared< std::packaged_task<return_type()> >(
        std::bind(std::forward<F>(f), std::forward<Args>(args)...)
        );

    std::future<return_type> res = task->get_future();
    {
        std::unique_lock<std::mutex> lock(queue_mutex_m);

        // don't allow enqueueing after stopping the pool
        if (stop_m)
            throw std::runtime_error("enqueue on stopped ThreadPool");

        std::thread::id id = std::this_thread::get_id();
        tasks_m.emplace([task](){ (*task)(); });
    }
    condition_m.notify_one();
    return res;
}

// the destructor joins all threads
inline ThreadPool::~ThreadPool()
{
    {
        std::unique_lock<std::mutex> lock(queue_mutex_m);
        stop_m = true;
    }
    condition_m.notify_all();
    for (std::thread &worker : workers_m)
        worker.join();
}

Answer 1

你的enqueue存储副本是不是要调用它们的参数？

如果您直接使用thread，那就是问题，您可以使用引用包装器 - 将std::ref(h)而不是h - 传递给线程构造函数，以便通过h作为参考。

我认为应该对你的ThreadPool做同样的事情。（如果这不起作用，则需要重新设计ThreadPool以使其正常工作）

Answer 2

问题详见于网址：

progress tag

解决方案是替换：

pool.enqueue(&MultiThreadedUpdateOfVector::setArray, h, begin, end)

由：

pool.enqueue(&MultiThreadedUpdateOfVector::setArray, std::ref(h), begin, end)

矢量的多线程更新

2 个答案: