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#include <memory>
 
struct Base {
  virtual void f();
};
 
struct Derived : Base {};
 
void f() {
  std::unique_ptr<Base> b = std::make_unique<Derivedunique<Derived>()>();
}

Compliant Solution

In this compliant solution, the destructor for Base has an explicitly declared virtual destructor, ensuring that the polymorphic delete operation results in well-defined behavior.

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Note that if Derived were not marked as final, then delete this could actually reference a subclass of Derived, violating this rule.

OOP52-CPP:EX1: Deleting a polymorphic object without a virtual destructor is permitted if its base class has a destroying operator delete that will figure out the correct derived class's destructor to call by other means.

#include <new>

class Base {
  const int whichDerived;

protected:
  Base(int whichDerived) : whichDerived(whichDerived) {}

public:
  Base() : Base(0) {}
  void operator delete(Base *, std::destroying_delete_t);
};

struct Derived1 final : Base {
  Derived1() : Base(1) {}
};

struct Derived2 final : Base {
  Derived2() : Base(2) {}
};

void Base::operator delete(Base *b, std::destroying_delete_t) {
  switch (b->whichDerived) {
  case 0:
    b->~Base();
    break;
  case 1:
    static_cast<Derived1 *>(b)->~Derived1();
    break;
  case 2:
    static_cast<Derived2 *>(b)->~Derived2();
  }
  ::operator delete(b);
}

void f() {
  Base *b = new Derived1();
  // ...
  delete b;
}

Risk Assessment

Attempting to destruct a polymorphic object that does not have a virtual destructor declared results in undefined behavior. In practice, potential consequences include abnormal program termination and memory leaks. 

Automated Detection

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