按索引访问可变参数模板中的类型

1 c++ decltype boost-any variadic-templates c++11

我想通过索引获得可变参数模板中的类型.索引被指定为模板参数.我设法找到一个有效的'hack',但我相信它不符合可变参数模板编程的精神.此外,它使用额外的内存.

这是代码和一些解释:

template <typename... InputPortTypes>
class PipelineReceiver
{

protected:

    // This tuple is used for storing types only
    // Hence, I would like to get rid of it, but I am not sure how.
    std::tuple<
    std::function<std::unique_ptr<InputPortTypes> (int)>...
    > InputPortsTuple;

    // This vector is used for storing the actual objects
    // This is needed to be able to access/change its elements
    // during run time later on.
    // The vector is used for storage of function pointers (i.e. of type std::function)
    // that represent methods of another object upstream the pipeline.
    std::vector<boost::any> InputPortsVector;

public:

    PipelineReceiver()
        {
            // create an empty vector of the required size
            InputPortsVector.resize(sizeof...(InputPortTypes));
        }

    void connectPorts(int InputPortIndex, boost::any c_OutputPort)
        {
            // connect ports
            InputPortsVector[InputPortIndex] = c_OutputPort;
        }

     // this function needs to be modified to avoid using InputPortsTuple
    template<int N>
    void getInputPortValue(void)
        {
            std::cout <<
                *boost::any_cast<decltype(std::get<N>(this -> InputPortsTuple))>(
                    InputPortsVector[N]
                    )(0) <<
                std::endl;
        }

};
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我想删除该对象InputPortsTuple,并用某种形式的递归过程替换它来推断类型getInputPortValue.

理想情况下,我想N成为一个动态参数而不是模板参数.但是,我不确定这是否可行.

Win*_*ute 5

你可以简单地滥用std::tuple_element:

typename std::tuple_element<N, std::tuple<InputPortTypes...>>::type
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注意:如果你可以使用C++ 14,

std::tuple_element_t<N, std::tuple<InputPortTypes...>>
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做同样的事情是一种更好的方式.但是,并非所有常见的编译器都知道它.