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且构网 - 分享程序员编程开发的那些事

有没有办法让函数返回一个类型名?

更新时间:2021-10-26 01:40:41

您可以使用 模板元编程 实现目标的技巧.

You can use template metaprogramming techniques to accomplish your goal.

template <typename T1, typename T2> struct TypeSelector
{
   using type = double;
};

template <typename T1> struct TypeSelector<T1, int>
{
   using type = int;
};

template <typename T2> struct TypeSelector<int, T2>
{
   using type = int;
};

template <> struct TypeSelector<int, int>
{
   using type = int;
};

然后,使用:

int main()
{
    int a = 3, b = 2;
    using type1 = TypeSelector<decltype(a), decltype(b)>::type;
    std::cout << (static_cast<type1>(a) / static_cast<type1>(b)) << std::endl;

   float c = 4.5f;
   using type2 = TypeSelector<decltype(a), decltype(c)>::type;
   std::cout << (static_cast<type2>(a) / static_cast<type2>(c)) << std::endl;

   using type3 = TypeSelector<decltype(c), decltype(a)>::type;
   std::cout << (static_cast<type3>(c) / static_cast<type3>(a)) << std::endl;

}