c++ - 解包方法参数的元组

Question

我在将元组扩展为其内容时遇到语法问题。

我的工作代码:

class Example
{
    public:
        static void Go( int i, float f)
        {
            std::cout << "p1: " << i << std::endl;
            std::cout << "p2: " << f << std::endl;
        }

        template <typename T, size_t ... I>
            static void Do( T parm )
        {
            Go( std::get<I>( parm)...);
        }
};

int main()
{
    using X = std::tuple<int, float>;
    Example::Do<X,0,1>( std::make_tuple( 1,2.2)) ;
}

但是我想用类似的东西来调用扩展

int main()
{
    using X = std::tuple<int, float>;
    using IDX = std::std::index_sequence_for<int, float>;
    Example::Do<X,IDX>( std::make_tuple( 1,2.2)) ;
}

所以我正在寻找类似的东西(无法编译......):

template <typename T, size_t ... I>
static void Do<T, std::index_sequence<I...>>(T parm)
   {
            Go( std::get<I>( parm)...);
   }

Answer 1

按值传递索引序列:

template <typename T, size_t... I>
static void Do(T parm, std::index_sequence<I...>)
{
    Go(std::get<I>(parm)...);
}

像这样调用方法:

Example::Do(std::make_tuple(1, 2.2), 
    std::index_sequence_for<int, float>{});