10

我正在开发一个事件库,但我遇到了可变参数模板的问题。

一切都很好,除了我不能将引用作为参数传递......

这是一个非常简化的示例,用于揭露我的问题。

struct DelayedSignal 
{   
    ~DelayedSignal ()
    { std::cout << "~DelayedSignal CLOSE" << std::endl; }

    template<class C, class... Args>
    DelayedSignal ( void(C::*func)(Args...), C& obj )
    { std::cout << "DelayedSignal INIT - 03 - pointer to method & pointer to class instance (Arg num: " << sizeof...(Args) << ")" << std::endl; }

    template<class C, class... Args>
    DelayedSignal ( void(C::*func)(Args...), C& obj, Args... args )
    {
        std::cout << "DelayedSignal INIT - 04 - pointer to method & pointer to class instance & arguments (Arg num: " << sizeof...(Args) << ")" << std::endl;
    }
};

template<class... ArgsBis>
struct DelayedSignal_DebugHelper 
{
    ~DelayedSignal_DebugHelper ()
    { std::cout << "~DelayedSignal_DebugHelper CLOSE" << std::endl; }

    template<class C, class... Args>
    DelayedSignal_DebugHelper ( void(C::*func)(Args...), C& obj )
    { std::cout << "DelayedSignal_DebugHelper INIT - 03 - pointer to method & pointer to class instance (Arg num: " << sizeof...(Args) << ")" << std::endl; }

    template<class C, class... Args>
    DelayedSignal_DebugHelper ( void(C::*func)(Args...), C& obj, ArgsBis... args ) // Need to use ArgsBis instead of Args to make it work
    {
        std::cout << "DelayedSignal_DebugHelper INIT - 04 - pointer to method & pointer to class instance & arguments (Arg num: " << sizeof...(Args) << ")" << std::endl;
    }
};


template < class Tr, class... Args >
struct Signal
{
    void fire ( Args... args ) { std::cout << "Signal::fire::" << sizeof...(Args) << std::endl; }
};

struct Klass {};


int main()
{
    std::string str1("Blop");   // Will be used as reference
    Klass k;                    // Will be used as reference

    Signal<void, Klass&> signal_01;
    Signal<void, std::string&> signal_02;

    std::cout << "====== DelayedSignal :: needed for production purpose ===============" << std::endl;

    // OK
    DelayedSignal test01(&Signal<void, std::string&>::fire, signal_02);
    // HERE IS THE PROBLEM
    //DelayedSignal test02(&Signal<void, std::string&>::fire, signal_02, str1);

    // OK
    DelayedSignal test03(&Signal<void, Klass&>::fire, signal_01);
    // HERE IS THE PROBLEM
    //DelayedSignal test04(&Signal<void, Klass&>::fire, signal_01, k);

    std::cout << "====== DelayedSignal_DebugHelper :: used only for debug purpose ======" << std::endl;

    // OK
    DelayedSignal_DebugHelper<std::string&> test05(&Signal<void, std::string&>::fire, signal_02);
    // OK
    DelayedSignal_DebugHelper<std::string&> test06(&Signal<void, std::string&>::fire, signal_02, str1);

    // OK
    DelayedSignal_DebugHelper<Klass&> test07(&Signal<void, Klass&>::fire, signal_01);
    // OK
    DelayedSignal_DebugHelper<Klass&> test08(&Signal<void, Klass&>::fire, signal_01, k);

    return 1;
}

当我将所有 DelayedSignal 实例注册到单个 std::list 实例中时,我想避免在类本身上使用模板,这就是我在构造函数上使用模板的原因。我也可以使用纯虚类作为所有 DelayedSignal 的基础,并将指向虚类的指针注册到 std::list 但我认为最好尽量减少虚方法的使用,我对这个问题真的很感兴趣......

正如您在此示例中看到的,如果 test02 和 test04 被激活,它们将返回错误。DelayedSignal_DebugHelper 几乎与 DelayedSignal 相同,除了它在最后一个构造函数上使用 ArgsBis(类模板参数)而不是 Args 模板(方法模板参数),否则它不起作用(与 DelayedSignal 一样)。Args 是被接受的,但尽管它们在同一个构造函数声明中,void(C::*func)(Args...)但它并不被接受。ArgsBis... args

据我所知,只要没有引用,没有引用(DelayedSignal test04(&Signal<void, Klass>::fire, signal_01, k);例如)或有多个参数(或没有)就没有问题。

有没有办法解决这个问题?

谢谢你。

4

3 回答 3

2

我正在使用 clang,它给出了一个非常棒的错误消息:

test.cpp:59:19: error: no matching constructor for initialization of 'DelayedSignal'
    DelayedSignal test02(&Signal<void, std::string&>::fire, signal_02, str1);
                  ^      ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
test.cpp:9:5: note: candidate constructor template not viable: requires 2 arguments, but 3 were provided
    DelayedSignal ( void(C::*func)(Args...), C& obj )
    ^
test.cpp:13:5: note: candidate template ignored: deduced conflicting types for parameter 'Args'
      (<std::__1::basic_string<char> &> vs. <std::__1::basic_string<char>>)
    DelayedSignal ( void(C::*func)(Args...), C& obj, Args... args )
    ^

编译器推断出有冲突的类型Args

  1. std::string&
  2. std::string

我相信解决此问题的最佳方法正是您使用DelayedSignal_DebugHelper.

于 2011-04-04T13:16:37.137 回答
1

正如您已接受答案,这只是一个补充。虽然我可能会忽略一些东西,但identity像下面这样的类模板似乎使您的代码可编译。
例如:

template<class T> struct identity { typedef T type; };

struct DelayedSignal
{
    ...
    template<class C, class... Args>
    DelayedSignal ( void(C::*func)(Args...), C& obj, typename identity<Args>::type... args )
    {
       ...
    }
};

这是对ideone的测试

于 2011-04-05T20:10:12.297 回答
1

Howard Hinnant 是对的......您的另一种可能性是在任何地方使用引用,例如:

#include <iostream>

struct DelayedSignal 
{   
    ~DelayedSignal ()
     { std::cout << "~DelayedSignal CLOSE" << std::endl; }

    template<class C, class... Args>
    DelayedSignal ( void(C::*func)(Args &...), C& obj )
    { std::cout << "DelayedSignal INIT - 03 - pointer to method & pointer to class instance (Arg num: " << sizeof...(Args) << ")" << std::endl; }

    template<class C, class... Args>
    DelayedSignal ( void(C::*func)(Args &...), C& obj, Args & ... args )
    {
        std::cout << "DelayedSignal INIT - 04 - pointer to method & pointer to class instance & arguments (Arg num: " << sizeof...(Args) << ")" << std::endl;
    }
};

template < class Tr, class... Args >
struct Signal
{
     void fire ( Args &... args ) { std::cout << "Signal::fire::" << sizeof...(Args) << std::endl; }
};

struct Klass {};

int main()
{
    std::string str1("Blop");   // Will be used as reference
    Klass k;                    // Will be used as reference

    Signal<void, Klass&> signal_01;
    Signal<void, std::string&> signal_02;

    std::cout << "====== DelayedSignal :: needed for production purpose ===============" << std::endl;

    // OK
    DelayedSignal test01(&Signal<void, std::string&>::fire, signal_02);
    // HERE IS THE PROBLEM
    DelayedSignal test02(&Signal<void, std::string&>::fire, signal_02, str1);

}
于 2011-04-04T17:16:40.920 回答