C++ TypeId to a constructor

Question

I am trying to create a function that checks if type of object equal the type of argument, and if it does, create new object. My class hierarchy:

class Copier: public Action class CopierX: public Copier

So my function is:

void Copier::checkequality(Action b) {
    if (typeid(*this) == typeid(b))
    {
        Action b = new T x; //here at 'T' i want new type type(this)
    }
}

Just i think it is needed to mention - this is mentioned to be a virtual function with some 'default value', so there's another question, will it work like that:

so in this CopierX, typeid(this) will return CopierX type right?, then, i want to create something that will work like Action b = new CopierX x;

I would like to say sorry if i wrote it badly, i am fairly new to C++, i used to work a lot on simple C

Maybe you want to [clone](https://stackoverflow.com/q/12902751/10077) objects? — Fred Larson, May 12 '20 at 19:20
You're comparing type of a pointer to the type of a class object. They will never be the same. — eerorika, May 12 '20 at 19:22

score 3 · Accepted Answer · answered May 12 '20 at 19:45

One way to solve this would be adding Action* copy() method to the Action class and implementing it within every descendant.

The other is using static polymorphism like this

template<typename T>
class Action {
public:
template<typename U>
void checkequality(Action<U>** b)
{
    if(std::is_same<T, U>::value) {
        *b = new T();
    }
}
};

class Copier : public Action<Copier>
{}

C++ TypeId to a constructor

1 Answers1