c++ - 沿枚举C返回变量类型

我需要它来做我正在做的套接字缓冲区(读和写...)，这比其他任何东西都更有用，所以我想知道是否有一种更好的方法可以使用当前的c++功能/标准？我敢肯定。

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:

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

感谢您的答复。

最佳答案

您可以使用良好的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);

如果您想沿着基于constexpr的路线走，您可以执行以下操作:

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);

但是我发现这并不优于传统方法。

关于c++ - 沿枚举C返回变量类型，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/49405975/