必须将std：：std::vector<std::thread>中的线程连接两次，以避免线程dtor终止

Question

我编写了一个并行程序来找出两个求和问题：

#include <iostream>
#include <vector>
#include <algorithm>
#include <utility>
#include <functional>

#include <thread>
#include <atomic>
#include <future>

using std::vector;
class Solution
{
private:
    using val_t = std::pair<int, int>;
    using Container = vector<val_t>;
    using It = typename Container::iterator;
    using size_t = typename Container::size_type;

    static bool cmp(const val_t &x, const val_t &y) noexcept
    {
        return x.first < y.first;
    }

    // Precondition: data.size() == 0, nums.size() != 0
    static void make_data(Container &data, const vector<int> &nums)
    {
        data.reserve(nums.size());

        int i = 0;
        for (auto &each: nums)
            data.emplace_back(each, i++);

        std::sort(data.begin(), data.end(), &cmp);
    }

    // launch_threads will launch threads in parallel.
    // It will call F with an int(thread id for accessing data) and std::forward<Args>(args)...
    template <class F>
    static void launch_threads(std::reference_wrapper<vector<std::thread>> pool, int thread_cnt, F &&f)
    {
        auto launch_other_threads = [=, f = std::forward<F>(f)]()
        {
            // This is thread 0!
            for (int i = 1; i != thread_cnt; ++i)
                pool.get().emplace_back(f, i);

            return std::invoke(f, 0);
        };

        pool.get().reserve(thread_cnt);
        pool.get().emplace_back(launch_other_threads);
    }

    // number of threads this process will ever have(including the main thread)
    int total_threads;
    // Thread pool and how much data each will process
    vector<std::thread> pool;
    size_t n;

    // The data for processing
    int target;
    Container data;
    // is_ready indicates whether the data is ready
    std::shared_future<void> is_ready;

    // Where result will be put. 
    // Since there is only one result, no atomic variable is needed
    vector<int> ret;
    // Synchronization of the event to terminate other running threads
    std::atomic_bool finished;

    void findtwoSum_impl(int thread_id, It beg, It end) noexcept
    {
        for (; beg != end && !finished.load(std::memory_order_acquire); ++beg) {
            int val = target - beg->first;
            auto it = std::lower_bound(data.begin(), data.end(), val_t{val, 0}, &cmp);

            if (it != data.end() && it->first == val &&
                (it->second != beg->second || (++it)->first == val)) {

                finished.store(true, std::memory_order_release);
                ret = {beg->second, it->second};
                break;
            }
        }
    }

    void findtwoSum(int thread_id) noexcept
    {
        std::shared_future<void>{is_ready}.wait();

        // Calculate the data that this thread will process
        auto beg = data.begin() + n * thread_id;
        auto end = beg + n;

        return findtwoSum_impl(thread_id, beg, end);
    }

    // thread_cnt must > 0
    void launch_threads_and_prepare_data(const vector<int> &nums, int thread_cnt)
    {
        // Initial the notification mechanism
        std::promise<void> promise;
        is_ready = promise.get_future().share();

        // Launch threads
        launch_threads(pool, thread_cnt, [this](int id) noexcept {
            return findtwoSum(id);
        });

        // Prepare data
        make_data(data, nums);

        n = data.size() / total_threads;

        // Notify threads that the data is ready
        promise.set_value();
    }

    // do_last_thread_cleanup process the remaining data and join threads
    void do_last_thread_cleanup()
    {
        // The main thred is the last thread.
        int thread_id = total_threads - 1;

        // findtwoSum_impl returns if any thread find the result.
        findtwoSum_impl(thread_id, data.begin() + n * thread_id, data.end());

        // The original join loop
        // Wait for other threads to finish.
        for (auto &thread: pool)
            thread.join();

        // Clear containers
        pool.clear();
        data.clear();
    }

public:
    vector<int> twoSum(const vector<int> &nums, int target)
    {
        // Initialize class variables
        // I know that total_threads should be much lower depending on the size of input
        total_threads = 8;

        this->target = target;
        ret.reserve(2);
        finished.store(false, std::memory_order_release);

        // Initialize class variable pool, n, data and is_ready
        launch_threads_and_prepare_data(nums, total_threads - 1);

        do_last_thread_cleanup();

        return std::move(ret);
    }
};

int main()
{
    Solution s;
    s.twoSum({3, 2, 4}, 6);
    return 0;
}

我用clang++-8 -std=c++17 -O1 -g -fsanitizer=address -lpthread -o debug.out编译了它，当我运行./debug.out时，它毫无例外地被终止。

我尝试通过添加std::cerr << "!@@@" << std::endl;来调试它，在join()在std::vector<std::thread>中的std::threads之后，在Solution s超出范围之前；结果显示，pool.clear()是导致这种情况的代码。

我完全糊涂了，因为我在打电话给join之前就做了pool.clear()。为了发现问题，我将原始join()的代码修改为以下代码：

// The first loop
for (auto &thread: pool) {
    thread.join();
    std::cerr << " 1" << thread.get_id() << " is joinable? " << thread.joinable() << std::endl;
}

// The second loop
for (auto &thread: pool)
    if (thread.joinable())
        thread.join();
// The third loop
for (auto &thread: pool)
    std::cerr << thread.get_id() << " is joinable? " << thread.joinable() << std::endl;

而且，让我再次感到惊讶的是，我发现了join的第一个循环--线程根本不工作：

 1thread::id of a non-executing thread is joinable? 0
thread::id of a non-executing thread is joinable? 0
thread::id of a non-executing thread is joinable? 0
thread::id of a non-executing thread is joinable? 0
thread::id of a non-executing thread is joinable? 0
thread::id of a non-executing thread is joinable? 0
thread::id of a non-executing thread is joinable? 0
thread::id of a non-executing thread is joinable? 0

我认为我编写的第一个循环有问题，所以我注释掉了，然后再次运行它：

thread::id of a non-executing thread is joinable? 0
140634635626240 is joinable? 1
140634627233536 is joinable? 1
140634618840832 is joinable? 1
140634610448128 is joinable? 1
140634602055424 is joinable? 1
140634593662720 is joinable? 1
terminate called without an active exception

我完全糊涂了，也不知道如何解决这个问题。

Answer 1

您的bug

您的线程从n = 0开始，因为data.size()比total_threads小，整数除法n = data.size() / total_threads;将给出零。

void findtwoSum(int thread_id) noexcept
{
    std::shared_future<void>{is_ready}.wait();

    // Calculate the data that this thread will process
    auto beg = data.begin() + n * thread_id;
    auto end = beg + n;

    return findtwoSum_impl(thread_id, beg, end);
}

您有以下beg == end == data.begin()。此时没有线程执行任何计算并退出。

thread::id of a non-executing thread is joinable? 0

这是因为线程已经完成。

140634635626240 is joinable? 1

这是线程仍在运行\runnable的时候。

由于线程的调度是完全随机的，所以输出总是会变化的。这是正常的，即使您的代码中没有任何错误。

请注意：在启动线程之前设置所有内容:这避免了对std::shared_future<void> is_ready;的尴尬依赖