通过模板参数给定其长度，在编译时生成相同类型的std :: tuple [英] Produce std::tuple of same type in compile time given its length by a template argument

问题描述

在c ++中，我如何实现带有int模板参数(指示元组长度)的函数并生成具有该长度的std :: tuple?

In c++, how can I implement a function with an int template argument indicating the tuple length and produce a std::tuple with that length?

例如

func<2>() returns std::tuple<int, int>();
func<5>() returns std::tuple<int, int, int, int, int>().

推荐答案

这是带有别名模板的递归解决方案，可在C ++ 11中实现:

Here is a recursive solution with alias template and it's implementable in C++11:

template <size_t I,typename T> 
struct tuple_n{
    template< typename...Args> using type = typename tuple_n<I-1, T>::template type<T, Args...>;
};

template <typename T> 
struct tuple_n<0, T> {
    template<typename...Args> using type = std::tuple<Args...>;   
};
template <size_t I,typename T>  using tuple_of = typename tuple_n<I,T>::template type<>;

例如，如果我们想要"tuple of 3 doubles"，我们可以写:

For example if we want "tuple of 3 doubles" we can write:

tuple_of<3, double> t;

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