Question

If I have an interface and many classes that implement this interface, can I now pass as argument only the type of the class and not the object?

something like this:

Interface *creatClass(class : Interface){
    return new class();
}

EDIT:

template <class T>
IFrame *creatClass(){
    return new T();
}

void dfg(){
    IFrame *lol = creatClass<Button>();
}

error C3206: 'creatClass' : invalid template argument for 'Dist_Frame', missing template argument list on class template 'Button'

PS. Button inherits IFrame

Answer 1

It’s called “Template“.

template <class T>
T *creatClass(){
    return new T();
}

You’ll have to call it this way:

class InterfaceImpl : public Interface {...};
// ...
InterfaceImpl *newImpl = creatClass<InterfaceImpl>();

edited

You can also try this, to ensure you only create instances of Interface:

template <class T>
Interface *creatClass(){
    return new T();
}

edit to your edit

I tried this test code:

#include <iostream>
using namespace std;

class IFrame{
   public:
      virtual void Create() =0;
};

class Button : public IFrame{
    public:

       virtual void Create(){ cout << "In Button" << endl;};
};

template <class T>
IFrame *creatClass(){
    return new T();
}


int main()
{
    IFrame *lol = creatClass<Button>();
    lol->Create();
}

Works exactly as expected. You must have some other coding errors in your class definitions. Debug it, it has nothing to do with your original question.