Retrieving a c++ class name programmatically

Question

I was wondering if it is possible in C++ to retrieve the name of a class in string form without having to hardcode it into a variable or a getter. I\'m aware that none of th

Answer 1

What about this,

Tested on Windows 10 using Visual Studio 2019 (v142).

#include <iostream>
#include <typeinfo>
#include <string>

/**
 @author    blongho
 @fn        template<typename Object> std::string classNameOf()

 @brief     Determine the class name of an object

 @tparam    Object  Type of the object.

 @returns   A name of the class
 @date      2019-09-06
 */

template<typename Object>
std::string classNameOf() {
    std::string name = typeid(Object).name(); //* user defined types gives "class Type"*\ 
    size_t spacePosition = name.find_first_of(" ");
    if (spacePosition != std::string::npos) {
        return name.substr(spacePosition + 1, name.length());
    }
    return name; // mostly primitive types
}


class Person {
private:
    /* data */
public:
    Person() {};
    ~Person() {};

};

class Data
{
private:
    /* data */
public:
    Data() {};
    ~Data() {};

};

struct Type {};

int main() {
    std::cout << "Class name of Person() is \"" << classNameOf<Person>() << "\"\n";
    std::cout << "Class name of Data() is \"" << classNameOf<Data>() << "\"\n";
    std::cout << "Class name of Type() is \"" << classNameOf<Type>() << "\"\n";
    std::cout << "Class name of double is \"" << classNameOf<double>() << "\"\n";
    std::cout << "Class name of std::string is \"" << classNameOf<std::string>() << "\"\n";
    std::cout << "Class name of int is \"" << classNameOf<int>() << "\"\n";
    std::cout << "Class name of float is \"" << classNameOf<float>() << "\"\n";
    std::cout << "Class name of char is \"" << classNameOf<char>() << "\"\n";
    return 0;
}

---

Output

Class name of Person() is "Person"
Class name of Data() is "Data"
Class name of Type() is "Type"
Class name of double is "double"
Class name of std::string is "std::basic_string<char,struct std::char_traits<char>,class std::allocator<char> >"
Class name of int is "int"
Class name of float is "float"
Class name of char is "char"

In Ubuntu 18.04,

g++ -o test src/main.cpp
./test
Class name of Person() is "6Person"
Class name of Data() is "4Data"
Class name of Type() is "4Type"
Class name of double is "d"
Class name of std::string is "NSt7__cxx1112basic_stringIcSt11char_traitsIcESaIcEEE"
Class name of int is "i"
Class name of float is "f"
Class name of char is "c"

Answer 2

If you just want to check if it's certain class, then

typeid(obj) == typeid(CSubClass)

will always work regardless of the implementations.

Otherwise, a convenient way is to declare:

virtual const char* classname() { return "CMyClass";}

and implement per subclass.

Answer 3

You can use typeid:

#include <typeinfo>

std::cout << typeid(obj).name() << "\n";

However, the type name isn't standardided and may differ between different compilers (or even different versions of the same compiler), and it is generally not human readable because it is mangled.

On GCC and clang (with libstdc++ and libc++), you can demangle names using the __cxa_demangle function (on MSVC demangling does not seem necessary):

#include <cxxabi.h>
#include <cstdlib>
#include <memory>
#include <string>

std::string demangle(char const* mangled) {
    auto ptr = std::unique_ptr<char, decltype(& std::free)>{
        abi::__cxa_demangle(mangled, nullptr, nullptr, nullptr),
        std::free
    };
    return {ptr.get()};
}

This will still not necessarily be a readable name — for instance, std::string is a type name for the actual type, and its complete type name in the current libstdc++ is std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >; by contrast, in the current libc++ it’s std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> >. “Prettifying” type aliases is unfortunately not trivial.

Answer 4

The typeid(obj).name() thing always gives the type of the variable as it was declared, not the actual type (class) of the object. If the variable obj is assigned to an instance of a subclass of the class that obj was declared as, typeid doesn't reveal that, unfortunately.

Answer 5

With C++17, and a third-party library, you can now obtain the name of a class like

#include <iostream>
#include "nameof.hpp"

namespace test {
    class Object {};
}

int main() {
    constexpr auto obj_name = nameof::nameof_type<test::Object>();
    std::cout << obj_name << std::endl;
    // this prints "test::Object"
}

This uses only compile-time information, so it can be constexpr. Note that it’s not portable; for example Intel’s compiler isn’t supported.