枚举值的C ++十进制类型

Question

我有一个使用旧式枚举来标记类型的现有代码库：

enum value_type {
    FLOAT = 1,
    DOUBLE = 2,
    INT = 3
};

然后，我想std::vector<T>基于枚举值实例化-即类似：

auto make_vector( value_type enum_value ) -> decltype(std::vector<decltype( enum_value )>) {
   ...
}

当然，这decltype(enum_value)是行不通的-我想要像道德上这样的东西：

if (enum_value == FLOAT)
   return decltype(double());

if (enum_value == DOUBLE)
   return decltype(float());   

...

这样完全有可能-不用求助于if (enum_value == )编程风格吗？

Answer 1

what I would want something like the moral equivalent of [...] is something like this at all possible - without resorting to the if (enum_value == ) style of programming?

It's not possible if you pass enum_value as argument (run-time known) to make_array().

C++ is a static typed language, so the compiler must decide compile-time the return type of the function, so the compiler must known compile-time the enum_value.

To solve this problem, you can pass the enum_value as template parameter.

About the enum/type conversion, it seems to me that you need something as follows

template <value_type>
struct getType;

template <> struct getType<FLOAT>  { using type = float; };
template <> struct getType<DOUBLE> { using type = double; };
template <> struct getType<INT>    { using type = int; };

Now you can write a make_vector() function as follows

template <value_type enum_value>
auto make_vector ()
   -> std::vector<typename getType<enum_value>::type>
 { return {}; }