在C ++中创建一个模板结构或从Integer键到变量类型的映射

Question

The problem I am facing is the following I have some data stored in an array of unknown type array . However there is function array_getDataType() which given the array returns an integer for instance UBIT8 where UBIT8 is a constant defined somewhere and this basically means that the type of the elements stored in the array is unsigned char . I was wondering if there was some way to create a map between these defined constants and actual types, something that would take in "BIT8" and return "unsigned char". I know there is a way to do the opposite with templates in the following way.

template< typename T >
struct type2Int
{
     enum { id = 0 }; 
};
template<> struct type2Int<unsigned char>  { enum { id = UBIT8 }; };

and this can later be used as

type2Int<unsigned char> type;

then

type.id 

would be whatever is the definition of UBIT8 I was wondering how to do the opposite.

Answer 1

I was wondering how to do the opposite.

In a similar way, using template specialization.

By example

#include <iostream>

template <std::size_t>
struct int2type;
// { using type = void; }; ???

template <>
struct int2type<0U>
 { using type = char; };

template <>
struct int2type<1U>
 { using type = int; };

template <>
struct int2type<2U>
 { using type = long; };

int main()
 {
   static_assert(std::is_same<typename int2type<0>::type, char>::value, "!");
   static_assert(std::is_same<typename int2type<1>::type, int>::value, "!");
   static_assert(std::is_same<typename int2type<2>::type, long>::value, "!");
 }

If you can't use C++11 (so no using ), you can use the good-old typedef

template <std::size_t>
struct int2type;

template <>
struct int2type<0U>
 { typedef char type; };

// ...