c++ - 使用 std::vector> 和 std::async 启动几个线程时的 Abort()

Question

我使用 4 个线程使用 thread_local 创建一些对象内存池。

我正在使用 std::vector<std::future<int>>和 std::async(std::launch::async, function);调度线程和std::for_each与 t.get取回它们的值(value)。这是代码:

struct GameObject
{
    int x_, y_, z_;
    int m_cost;

    GameObject() = default;
    GameObject(int x, int y, int z, int cost)
        : x_(x), y_(y), z_(z), m_cost(cost)
    {}
};

struct Elf : GameObject
{
    Elf() = default;
    Elf(int x, int y, int z, int cost)
        : GameObject(x, y, z, cost)
    {
        std::cout << "Elf created" << '\n';
    }

    ~Elf() noexcept
    {
        std::cout << "Elf destroyed" << '\n';
    }
    std::string m_cry = "\nA hymn for Gandalf\n";
};

struct Dwarf : GameObject
{
    Dwarf() = default;
    Dwarf(int x, int y, int z, int cost)
        : GameObject(x, y, z, cost)
    {
        std::cout << "dwarf created" << '\n';
    }

    ~Dwarf() noexcept
    {
        std::cout << "dwarf destroyed" << '\n';
    }
    std::string m_cry = "\nFind more cheer in a graveyard\n";
};


int elvenFunc()
{
    thread_local ObjectPool<Elf> elvenPool{ 229 };
    for (int i = 0; i < elvenPool.getSize(); ++i)
    {
        Elf* elf = elvenPool.construct(i, i + 1, i + 2, 100);
        std::cout << elf->m_cry << '\n';
        elvenPool.destroy(elf);
    }

    thread_local std::promise<int> pr;
    pr.set_value(rand());

    return 1024;
}

int dwarvenFunc()
{
    thread_local ObjectPool<Dwarf> dwarvenPool{ 256 };
    for (int i = 0; i < dwarvenPool.getSize(); ++i)
    {
        Dwarf* dwarf = dwarvenPool.construct(i - 1, i - 2, i - 3, 100);
        std::cout << dwarf->m_cry << '\n';
        dwarvenPool.destroy(dwarf);
    }

    thread_local std::promise<int> pr;
    pr.set_value(rand());

    return 2048;
}


int main()
{
    std::ios_base::sync_with_stdio(false);
    srand(time(0));

    std::vector<std::future<int>> vec{ 4 };
    vec.emplace_back(std::async(std::launch::async, elvenFunc));
    vec.emplace_back(std::async(std::launch::async, elvenFunc));
    vec.emplace_back(std::async(std::launch::async, dwarvenFunc));
    vec.emplace_back(std::async(std::launch::async, dwarvenFunc));

    int term = 0;
    try
    {
        std::for_each(std::execution::par, vec.begin(), vec.end(), [&term](std::future<int>& t)
        {
            auto ret = t.get();
            std::cout << "thread brought me " << ret << '\n';
            term += ret;
        });
    }
    catch (const std::exception& ex)
    {
        std::cout << ex.what() << '\n';
    }

    std::cout << "Final word = " << term << '\n';
}

(construct 和 destroy 在内部调用 allocate 和 deallocate。)我从终端得到了很多预期的输出，但在 abort 的某个地方被调用并且程序无法正常完成。我不知道为什么。我相信 t.get()调用以 std::async 启动的线程会自动调用 .join太对了吧？

使用 C++17 和 Visual Studio 2017。我做错了什么？

Answer 1

你有未定义的行为。您在无效的 future 调用 get。您的 future vector 的前 4 个项目为空 future。只有当 future::valid 返回 true 时，才能调用 get。

这行你怎么看

std::vector<std::future<int>> vec{ 4 };

是吗？

Default constructor. Constructs a std::future with no shared state. After construction, valid() == false.

和here您可以阅读调用 future::get 时发生的情况:

The behavior is undefined if valid() is false before the call to this function.