why type name is not allowed?-CodePudding

I want that when passing a certain enumeration property, the corresponding operator with type checking is output, here is a simple code example

#include <iostream>
#include <typeinfo>
//#include <memory>

enum MessageType : uint16_t
{
    Display,
    Warning,
    Error
};

#define MESSAGE(_MessageType, _Text)\
    if((_MessageType == Display) && (typeid(_MessageType).name() == MessageType))\
    {\
        std::cout<<"Display: "<<_Text;\
    }\
    else if((_MessageType == Warning) && (typeid(_MessageType).name() == MessageType))\
    {\
        std::cout<<"Warning: "<<_Text;\
    }\
    else if ((_MessageType == Warning) && (typeid(_MessageType).name() == MessageType))\
    {\
        std::cout<<"Error: "<<_Text;\
    }\
    else\
        return

int main()
{
    MESSAGE(Display, "Text!"); // type name is not allowed
    return 0;
}

CodePudding user response：

std::type_info::name returns a c-string. MessageType is not a string, its the name of a type. You can compare the string returned from typeid(_MessageType).name() to the string returned from typeid(MessageType).name().

However, identifiers starting with leading _ followed by capital letter are reseved. If you use them your code has undefined behavior. Moreover main is not void you must return an int.

And last but not least, you don't need typeid when you can use constexpr if and std::is_same. A simple function instead of macro would also make it much easier to get a compiler error when the type does not match. Currently your check for the right type is made at runtime. I see no reason to use a macro here, the same could be achieved without one.