我正在处理一个事件库,而我正在面对Variadic模板的问题.
所有的工作非常好,除了我不能传递引用作为参数的事实…
这是一个非常简化的例子,用来揭示我的问题.
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,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,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,signal_02,str1);
// OK
DelayedSignal test03(&Signal<void,Klass&>::fire,signal_01);
// HERE IS THE PROBLEM
//DelayedSignal test04(&Signal<void,signal_01,k);
std::cout << "====== DelayedSignal_DebugHelper :: used only for debug purpose ======" << std::endl;
// OK
DelayedSignal_DebugHelper<std::string&> test05(&Signal<void,signal_02);
// OK
DelayedSignal_DebugHelper<std::string&> test06(&Signal<void,str1);
// OK
DelayedSignal_DebugHelper<Klass&> test07(&Signal<void,signal_01);
// OK
DelayedSignal_DebugHelper<Klass&> test08(&Signal<void,k);
return 1;
}
当我将所有DelayedSignal实例注册到单个std :: list实例中时,我想避免在类本身使用模板,这就是为什么我在构造函数上使用模板.我也可以使用一个纯虚拟类作为所有DelayedSignal的基础,并将虚拟类的注册注册到std :: list中,但我认为最好最小化虚拟方法的使用,我真的很感兴趣这个问题？
正如你在这个例子中看到的那样,如果test02和test04被激活,返回错误.
DelayedSignal_DebugHelper与DelayedSignal几乎相同,除了在最后一个构造函数(而不是Args模板(方法模板参数))中使用ArgsBis(一个类模板参数),否则它不起作用(与DelayedSignal一样). Args被接受在void(C :: * func)(Args …)上,但不是与ArgsBis … args排除它们在同一个构造函数声明中的事实.
据我所知,只要没有参考(DelayedSignal test04(& Signal< void,Klass> :: fire,k)),或者有多个参数(或没有)引用.
有没有办法解决这个问题？
谢谢.
解决方法
Howard Hinnant是对的…另一种可能性是使用任何地方的引用,例如：
#include <iostream>
struct DelayedSignal
{
~DelayedSignal ()
{ std::cout << "~DelayedSignal CLOSE" << std::endl; }
template<class C,class... Args>
DelayedSignal ( void(C::*func)(Args &...),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,signal_02);
// HERE IS THE PROBLEM
DelayedSignal test02(&Signal<void,str1);
}
