C++ deleting a derived object via base pointer without a virtual destructor: what sequence prints? #include class CuriousTabBase { public: CuriousTabBase() { cout << "Base OK. "; } ~CuriousTabBase() { cout << "Base DEL. "; } }; class CuriousTabDerived : public CuriousTabBase { public: CuriousTabDerived() { cout << "Derived OK. "; } ~CuriousTabDerived() { cout << "Derived DEL. "; } }; int main() { CuriousTabBase basePtr = new CuriousTabDerived(); delete basePtr; // base dtor is non-virtual return 0; }

Introduction / Context: This classic test examines destructor polymorphism. Deleting a derived object through a base pointer whose destructor is not virtual leads to only the base destructor being invoked, which is undefined behavior in standard C++ but typically manifests as the derived destructor not running (a resource leak risk).

Given Data / Assumptions:

• CuriousTabBase destructor is non-virtual.
• Object is allocated as CuriousTabDerived but referenced via CuriousTabBase.
• delete basePtr; is used.

Concept / Approach: Construction of a derived object prints base message first, then derived message. On deletion through a base pointer without a virtual destructor, only the base destructor is called. The derived destructor is skipped, which is incorrect behavior but matches many legacy outputs.

Step-by-Step Solution:

Verification / Alternative check: Making ~CuriousTabBase() virtual would produce: Base OK. Derived OK. Derived DEL. Base DEL.

Why Other Options Are Wrong:

• Any option including Derived DEL on delete is incorrect without a virtual base destructor.
• Missing Base DEL omits the observed delete behavior.

Common Pitfalls: Forgetting to declare virtual destructors in base classes intended for polymorphic deletion; overlooking UB concerns despite predictable prints in many toolchains.

Final Answer: Base OK. Derived OK. Base DEL.