c++ - 沿枚举 C++ 返回变量类型

Question

我需要它用于我正在制作的套接字缓冲区(读取和写入......)，这比其他任何东西都更像是一个技巧所以我想知道是否有更好的方法来使用当前的C++ 特性/标准？我确定。

enum e_Types
{
    Type_64 ,
    Type_32 ,
    Type_16 ,
    Type_8 ,
    Type_ALL
};

template<e_Types Type> constexpr auto _GetVarType()
{
    if constexpr( Type == Type_8 )
        return ( unsigned char* ) 0;
    if constexpr( Type == Type_16 )
        return ( unsigned short* ) 0;
    if constexpr( Type == Type_32 )
        return ( unsigned long* ) 0;
    if constexpr( Type == Type_64 )
        return ( unsigned long long* ) 0;
}

#define GetVarType(type) decltype( _GetVarType<type>() )

例子:

GetVarType( Type_64 ) p64 = malloc(0x1000);

感谢回复。

Answer 1

您可以使用旧的显式模板类特化:

template <e_Types X> struct type_of;
template <>          struct type_of<Type_64> { using type = unsigned long long; };
template <>          struct type_of<Type_32> { using type = unsigned long; };
template <>          struct type_of<Type_16> { using type = unsigned short; };
template <>          struct type_of<Type_8>  { using type = unsigned char; };

template <e_Types X> 
using type_of_t = typename type_of<X>::type;

用法:

type_of_t<Type_64>* p64 = malloc(0x1000);

---

如果你想走基于 constexpr 的路线，你可以这样做:

template <typename T> 
struct type_wrapper 
{ 
    using type = T; 
};

template <typename T> 
inline constexpr type_wrapper<T> t{};

template <e_Types X> inline constexpr auto type_of          = t<void>;
template <>          inline constexpr auto type_of<Type_64> = t<unsigned long long>;
template <>          inline constexpr auto type_of<Type_32> = t<unsigned long>;
template <>          inline constexpr auto type_of<Type_16> = t<unsigned short>;
template <>          inline constexpr auto type_of<Type_8>  = t<unsigned char>;

用法:

typename decltype(type_of<Type_64>)::type* p64 = malloc(0x1000);

但我认为这并不优于更传统的方法。