使用“使用声明”扩展非类型模板参数包(模板可变参数编译时SignalSlot实现)

时间:2019-01-31 15:29:03

标签: c++ c++17 variadic-templates multiple-inheritance

有更好的标题建议吗?

在Qt中,有一个很好的信号和插槽功能。但是,它告诉您是否仅在运行时间(afc)内可以将特定信号连接到特定插槽。

打算:

  • 从模板创建一个包含“信号签名”(函数指针作为模板参数)的类,以允许将给定签名(传递参数的数量和类型)的“槽”仅连接到具有相似签名的“已定义”信号;

  • 必须易于使用。

现在的问题: 我在ISignalSlotMap类中收到“使用声明”的编译错误。 template multiple variadic inheritance with variadic argument types-在这里编译良好。

还有,有什么方法可以简化模板算法?

更新:第一个块可以在没有dll的情况下编译和运行

无需链接到DLL即可进行编译

#include <iostream>
#include <type_traits>
#include <forward_list>
#include <memory>

//template wrapper
template <typename...>
struct TW
{};


//template to get Class type from pointer
template <class ReturnType, class ... ArgTypes>
constexpr ReturnType ClassFromPointer(void(ReturnType::*)(ArgTypes...));


//template to get pack of arguments' types
template <class ReturnType, class ... ArgTypes>
constexpr TW<ArgTypes...> ArgTypesPackFromPointer(void(ReturnType::*)(ArgTypes...));

template <auto ptr>
using FuncClass = decltype(ClassFromPointer(ptr));

template <auto ptr>
using FuncPack = decltype(ArgTypesPackFromPointer(ptr));


template <class ... ArgTypes>
struct Invoker
{
    virtual void Invoke(ArgTypes ... args) = 0;
};


template <class ClType, class ... ArgTypes>
class InvokerImpl : public Invoker<ArgTypes...>
{
    ClType *ptr_;
    void(ClType::*pFunc_)(ArgTypes...);

public:
    InvokerImpl(ClType* pObj, void(ClType::*pFunc)(ArgTypes...))
        : ptr_(pObj),
        pFunc_(pFunc)
    {}

    virtual void Invoke(ArgTypes ... args)
    {
        (ptr_->*pFunc_)(args...);
    }
};

template <class ClType, class ... ArgTypes>
Invoker<ArgTypes...>* CreateInvoker(ClType* pObj, void(ClType::*pFunc)(ArgTypes...))
{
    return new InvokerImpl<ClType, ArgTypes...>(pObj, pFunc);
}

template <class Pack>
class SlotContainerTranslated;

template <template <class ...> class Pack, class ... ArgTypes>
class SlotContainerTranslated<Pack<ArgTypes...>>
{
    typedef std::unique_ptr<Invoker<ArgTypes...>> pInvoker;
    std::forward_list<pInvoker> slots_;

public:
    void AddInvoker(Invoker<ArgTypes...>* pInv)
    {
        slots_.push_front(std::move(pInvoker(pInv)));
    }

    void DispatchSignal(ArgTypes ... args)
    {
        auto start = slots_.begin();
        while (start != slots_.end())
        {
            (*start)->Invoke(args...);
            ++start;
        }
    }
};

template <auto memfuncptr>
class ISlotContainer : SlotContainerTranslated<FuncPack<memfuncptr>>
{
public:
    using SlotContainerTranslated<FuncPack<memfuncptr>>::AddInvoker;
    using SlotContainerTranslated<FuncPack<memfuncptr>>::DispatchSignal;
};


template <auto ... memfuncPtrs>
class ISignalSlotMap : SlotContainerTranslated<FuncPack<memfuncPtrs>>...
{
public:
    //  using SlotContainerTranslated<FuncPack<memfuncPtrs>>::AddInvoker...;
    //  using SlotContainerTranslated<FuncPack<memfuncPtrs>>::DispatchSignal...;

};
////////////////////////////////////////////////////////////////////////

struct AlienSignals
{
    void MindControl() {};
    void MindControlPrint(int a, double b, int c, int d, const char* str) {};
    void MindControlAdvise(int i, bool b) {};
};



struct Alien
{
    static Alien* Invade();
    virtual ISlotContainer<&AlienSignals::MindControlAdvise>& AccessSignal() = 0;

    /*//this is what usage is expected to be like
    virtual ISignalSlotMap<&AlienSignals::MindControl,
        &AlienSignals::MindControlAdvise,
        &AlienSignals::MindControlPrint>& AccessSignalMap() = 0;
        */

    virtual ~Alien() = default;
};

class AlienImpl : public Alien
{
    std::unique_ptr<ISlotContainer<&AlienSignals::MindControlAdvise>> signalMindControlAdvise_
    { new ISlotContainer<&AlienSignals::MindControlAdvise> };

    // Inherited via Alien
    virtual ISlotContainer<&AlienSignals::MindControlAdvise>& AccessSignal() override
    {
        return *signalMindControlAdvise_;
    }

    virtual ~AlienImpl() = default;
};

Alien * Alien::Invade()
{
    return new AlienImpl;
}


struct Human
{
    int id = 0;

    Human(int i)
        : id(i)
    {}

    void Print()
    {
        std::cout << "Human: " << id << "! " << std::endl;
    }

    void mPrint(int a, double b, int c, int d, const char* str)
    {
        std::cout << "Human: " << id << "! " << a << " " << b << " " << c << " " << d << " " << str << std::endl;
    }

    void Advise(int i, bool b)
    {
        auto colour = b ? "red" : "blue";
        std::cout << "Human: " << id << "! I will take " << i << " of " << colour << " pills" << std::endl;
    }
};

template <auto memfuncptr>
constexpr auto GetType()
{
    return memfuncptr;
}

template <auto memfunc>
using PtrType = decltype(GetType<memfunc>());

int main()
{
    Human person1{ 1 }, person2{ 2 }, person3{ 3 };

    std::unique_ptr<Alien>alien{ Alien::Invade() };
    alien->AccessSignal().AddInvoker(CreateInvoker(&person1, &Human::Advise));
    alien->AccessSignal().AddInvoker(CreateInvoker(&person2, &Human::Advise));
    alien->AccessSignal().AddInvoker(CreateInvoker(&person3, &Human::Advise));
    alien->AccessSignal().DispatchSignal(42, false);

    return 0;
}

UPDATE2: 我发现问题出在扩展非类型模板参数包,因此“使用”可以工作。我仍然无法克服这个问题。

c++ non-type parameter pack expansion类似的问题,但有关功能。我也找不到带有继承的折叠表达式的任何用法。

有一个答案显示了一种有前途的方法:https://stackoverflow.com/a/53112843/9363996

但是有主要缺点。一种是使用模板函数来调用继承的函数。此示例可以编译并运行,但是:

  • 如果要为DLL编译接口,我不知道如何强制从模板生成方法;
  • 由于intellisense没有显示期望什么参数,而您必须显式指定函数指针,这非常不便。

示例2

#include <iostream>

template <class ...>
struct TW {};

template <class ClType, class ... ArgTypes>
constexpr ClType ClassType(void(ClType::*)(ArgTypes...));

template <class ClType, class ... ArgTypes>
constexpr TW<ArgTypes...> ArgsType(void(ClType::*)(ArgTypes...));

template <auto pFunc>
using class_trait = decltype(ClassType(pFunc));

template <auto pFunc>
using args_trait = decltype(ArgsType(pFunc));

template <class, class>
struct _func_trait;

template <class ClType, template <class...> class Pack, class ... ArgTypes>
struct _func_trait<ClType, Pack<ArgTypes...>>
{
    typedef void(ClType::*FuncPtr)(ArgTypes...);
    typedef ClType ClassType;
    typedef Pack<ArgTypes...> Args;
};

template <auto pFunc>
struct func_traits : public _func_trait<class_trait<pFunc>, args_trait<pFunc>>
{};


template <auto L, class Pack>
struct ClassImpl;

template <auto L, template <class ...> class Pack, class ... ArgTypes>
struct ClassImpl<L, Pack<ArgTypes...>>
{
    void invoke(ArgTypes ... args)
    {
        (std::cout << ... << args) << std::endl;
    }
};

template <auto L, auto ...R>
class My_class;

template <auto L>
class My_class<L> : public ClassImpl <L, args_trait<L>>
{

};

template <auto L, auto ... R>
class My_class : public My_class<L>, public My_class<R...>
{
public:

    template <auto T, class ... ArgTypes>
    void Invoke(ArgTypes... args)
    {
        My_class<T>::invoke(args...);
        return;
    }

};



struct Signals
{
    void func1(int a, double b) {}

    void func2(const char*, const char*) {}

    constexpr void func3(int a, double b, int c, bool d);
};


int main()
{

    Signals s;
    My_class<&Signals::func1, &Signals::func2, &Signals::func3> mSignls;
    mSignls.Invoke<&Signals::func1>(4, 6.31);
    mSignls.Invoke<&Signals::func2>("Invoking funcion:", "function 2");

    return 0;
}

1 个答案:

答案 0 :(得分:0)

最后,我想出了一个解决方案,它的使用是非常简单的,因为我想。

这是我的工作示例!

#include <tuple>
#include <iostream>

template <class ...>
struct TW {};

template <class ClType, class ... ArgTypes>
constexpr ClType ClassType(void(ClType::*)(ArgTypes...));

template <class ClType, class ... ArgTypes>
constexpr TW<ArgTypes...> ArgsType(void(ClType::*)(ArgTypes...));

template <auto pFunc>
using class_trait = decltype(ClassType(pFunc));

template <auto pFunc>
using args_trait = decltype(ArgsType(pFunc));

template <class, class>
struct _func_trait;

template <class ClType, template <class...> class Pack, class ... ArgTypes>
struct _func_trait<ClType, Pack<ArgTypes...>>
{
    typedef void(ClType::*FuncPtr)(ArgTypes...);
    typedef ClType ClassType;
    typedef Pack<ArgTypes...> Args;
};

template <auto pFunc>
struct func_traits : public _func_trait<class_trait<pFunc>, args_trait<pFunc>>
{};


template <auto L, class Pack = args_trait<L>>
struct ClassImpl;

template <auto L, template <class ...> class Pack, class ... ArgTypes>
struct ClassImpl<L, Pack<ArgTypes...>>
{
    void invoke(decltype(L), ArgTypes ... args)
    {
        (std::cout << ... << args) << std::endl;
    }
};



template <class ... Impls>
struct ISignalMap : protected Impls...
{
    using Impls::invoke...;
};

template <auto ... L>
struct SignalsMap
{
    //just to see the pointers' values
    static constexpr std::tuple<decltype(L)...> t{ std::make_tuple(L...) };
    ISignalMap<ClassImpl<L>...> Signals{};
};

struct Signals
{
    void func1(int a, double b) {}
    void func12(int a, double b) {}

    void func2(double a, double b, int c) {}

    constexpr void func3(const char*) {}
};


int main(void)
{
    auto& ref = SignalsMap<&Signals::func1, &Signals::func2, &Signals::func3>::t;

    //add SignalsMap as member to your class and pass the pointers to
    //methods you need to be signals
    SignalsMap<&Signals::func1, &Signals::func2, &Signals::func3> sm;

    //first parameter is a pointer to a signal you want to invoke
    sm.Signals.invoke(&Signals::func2, 4.8, 15.16, 23);
    sm.Signals.invoke(&Signals::func1, 23, 42.108);
    sm.Signals.invoke(&Signals::func12, 23, 42.108);

    sm.Signals.invoke(&Signals::func3, "Eat this!");

    return 0;
}