Deducing this causes issues if you inherit from the base class more than once, (it increases object size) also, this doesn't work since clone() isn't virtual and templates can't be virtual
cpp
struct D {};
struct A : D {};
struct B : D {
A a;
};
you would expect since 'B' inherits from an empty class and it only has one member the sizeof would be 1, but it is 2 since the base class D is repeated twice and must have unique address first in the inheritance in B and in the member A.
now if this uses crtp the base classes would be unique
cpp
template<class T>
struct D {};
struct A : D<A> {};
struct B : D<B> {
A a;
}; // sizeof(B) == 1
No. According to the standard, two different objects of the same type cannot live at the same address in any circumstance, even if the objects are empty.
3
u/robin-m 12d ago
With c++23 deducing this you can simplify your code to just:
c++23 class Shape { public: template <class Self> std::unique_ptr<Shape> clone(this Self& self) { return std::make_unique<Self>(self); } }; class Rectangle: public Shape { }; class Triangle: public Shape { };