Question

我想通过从C ++ 11包装std :: thread类来使用我自己的Thread实现，这样我就能处理我想要的异常。

这是我的包装类：

#include <Types.hpp>
#include <thread>
#include <exception>
#include <functional>

class Thread
{
    private:

        std::exception_ptr exceptionPtr;
        std::thread thread;

    public:

        using Id = std::thread::id;

        using NativeHandleType = std::thread::native_handle_type;

        Thread() noexcept = default;
        Thread(Thread &&t) noexcept :
            exceptionPtr(std::move(t.exceptionPtr)),
            thread(std::move(t.thread))
        {
        }

        Thread &operator =(Thread &&t) noexcept
        {
            exceptionPtr = std::move(t.exceptionPtr);
            thread = std::move(t.thread);
            return *this;
        }

        template<typename Callable, typename... Args>
        Thread(Callable &&f, Args &&... args) :
            exceptionPtr(nullptr),
            thread([&](Callable &&f, Args &&... args)
            {
                try
                {
                    std::once_flag flag;
                    std::call_once(flag, f, args...);
                }
                catch (...)
                {
                    exceptionPtr = std::current_exception();
                }

            }, f, args...)
        {
            if (exceptionPtr != nullptr)
            {
                 std::rethrow_exception(exceptionPtr);
            }
        }

        bool joinable() const noexcept
        {
            return thread.joinable();
        }

        void join()
        {
            thread.join();
        }

        void detach()
        {
            thread.detach();
        }

        Id getId() const noexcept
        {
            return thread.get_id();
        }

        NativeHandleType nativeHandle()
        {
            return thread.native_handle();
        }

        static uint32_t hardwareConcurrency() noexcept
        {
            return std::thread::hardware_concurrency();
        }

        static void wait(Time t)
        {
            std::this_thread::sleep_for(t);
        }
};

如果没有争论，它的效果非常好：

Thread([&]() {  /* do something */ }).detach();

...但如果我尝试传递可变参数：

Thread(&GUI::refreshTask, this, refreshDelay).detach();

...我在编译时遇到错误：

buildroot-2014.02 / output / host / usr / i586-buildroot-linux-uclibc / include / c ++ / 4.8.2 / functional：实例化＆＃39; struct std :: _ Bind_simple）（std :: chrono：：duration＆gt;）; Args = {CRH :: GUI const，std :: chrono :: duration＆gt;＆amp;}] :: __ lambda1（void（CRH :: GUI :: ）（std :: chrono :: duration＆gt;）， CRH :: GUI ，std :: chrono :: duration＆gt;）＆gt;＆＃39;： buildroot-2014.02 / output / host / usr / i586-buildroot-linux-uclibc / include / c ++ / 4.8.2 / thread：137：47：需要来自＆＃39; std :: thread :: thread（_Callable＆amp;＆amp; ，_Args＆amp;＆amp; ...）[with _Callable = CRH :: Thread :: Thread（Callable＆amp;＆amp;，Args＆amp;＆amp; ...）[with Callable = void（CRH :: GUI :: ）（std :: chrono :: duration＆gt;）; Args = {CRH :: GUI const，std :: chrono :: duration＆gt;＆amp;}] :: __ lambda1; _Args = {void（CRH :: GUI :: ＆amp;）（std :: chrono :: duration＆gt;），CRH :: GUI const＆amp;，std :: chrono :: duration＆gt;＆amp; ;}]＆＃39; /home/cyril/Documents/crh-2016/src/robot2/../core/Thread.hpp:72:30：来自＆CRH :: Thread :: Thread（Callable＆amp;＆amp;，Args＆amp;＆amp; amp; ; ...）[with Callable = void（CRH :: GUI :: ）（std :: chrono :: duration＆gt;）; Args = {CRH :: GUI const，std :: chrono :: duration＆gt;＆amp;}]＆＃39; src / core / GUI.cpp:90:57：从这里需要 buildroot-2014.02 / output / host / usr / i586-buildroot-linux-uclibc / include / c ++ / 4.8.2 / functional：1697：61：错误：没有命名的类型＆＃39; type＆＃39; in＆＃39; class std :: result_of）（std :: chrono :: duration＆gt;）; Args = {CRH :: GUI const，std :: chrono :: duration＆gt;＆amp;}] :: __ lambda1（void（CRH :: GUI :: ）（std :: chrono :: duration＆gt;）， CRH :: GUI ，std :: chrono :: duration＆gt;）＆gt;＆＃39; typedef typename result_of＆lt; _Callable（_ Args ...）＆gt; :: type result_type; ^ buildroot-2014.02 / output / host / usr / i586-buildroot-linux-uclibc / include / c ++ / 4.8.2 / functional：1727：9：错误：没有命名的类型＆＃39; type＆＃39; in＆＃39; class std :: result_of）（std :: chrono :: duration＆gt;）; Args = {CRH :: GUI const，std :: chrono :: duration＆gt;＆amp;}] :: __ lambda1（void（CRH :: GUI :: ）（std :: chrono :: duration＆gt;）， CRH :: GUI ，std :: chrono :: duration＆gt;）＆gt;＆＃39; _M_invoke（_Index_tuple＆LT; _Indices ...＆GT）

可能会更清楚......但对GCC来说要求太高了。

知道如何解决这个问题吗？

解决方案

#include <Types.hpp>
#include <thread>
#include <exception>
#include <functional>

class Thread
{
    private:

        std::exception_ptr exceptionPtr;
        std::thread thread;

    public:

        using Id = std::thread::id;

        using NativeHandleType = std::thread::native_handle_type;

        Thread() noexcept = default;
        Thread(Thread &&t) noexcept :
            exceptionPtr(std::move(t.exceptionPtr)),
            thread(std::move(t.thread))
        {
        }

        Thread &operator =(Thread &&t) noexcept
        {
            exceptionPtr = std::move(t.exceptionPtr);
            thread = std::move(t.thread);
            return *this;
        }

        template<typename Callable, typename... Args>
        Thread(Callable &&f, Args &&... args) :
            exceptionPtr(nullptr),
            thread([&](typename std::decay<Callable>::type &&f, typename std::decay<Args>::type &&... args)
            {
                try
                {
                    std::bind(f, args...)();
                }
                catch (...)
                {
                    exceptionPtr = std::current_exception();
                }

            }, std::forward<Callable>(f), std::forward<Args>(args)...)
        {
        }

        bool joinable() const noexcept
        {
            return thread.joinable();
        }

        void join()
        {
            thread.join();

            if (exceptionPtr != nullptr)
            {
                std::rethrow_exception(exceptionPtr);
            }
        }

        void detach()
        {
            thread.detach();
        }

        Id getId() const noexcept
        {
            return thread.get_id();
        }

        NativeHandleType nativeHandle()
        {
            return thread.native_handle();
        }

        static uint32_t hardwareConcurrency() noexcept
        {
            return std::thread::hardware_concurrency();
        }

        static void wait(Time t)
        {
            std::this_thread::sleep_for(t);
        }
};

Answer 1

Callable和Args是转发引用，因此模板参数推导可以使它们成为左值引用或普通类型，具体取决于参数表达式的值类别。

这意味着当您在lambda声明中重用推导出的类型时：

thread([&](Callable&& f, Args&&... args)

参考折叠发挥作用，对于左值参数refreshDelay，Args成为左值参考。

但是，std::thread存储它接收的参数的衰减副本，然后它从内部存储移动到实际处理程序，将存储的对象转换为xvalues。这就是错误告诉你的：处理程序不能用线程试图传入的参数调用。

相反，您可以按如下方式实现它：

template <typename Callable, typename... Args>
Thread(Callable&& f, Args&&... args)
    : exceptionPtr(nullptr)
    , thread([] (typename std::decay<Callable>::type&& f
               , typename std::decay<Args>::type&&... args)
            {
                // (...)
            }
            , std::forward<Callable>(f), std::forward<Args>(args)...)
{
    // (...)
}

如何将可变参数args传递给std :: thread？

1 个答案: