Synopsis

A public non-virtual destructor is defined in a class with virtual methods.

Enabled by default

Yes

Severity/Certainty

Medium/High

Full description

A public non-virtual destructor is defined in a class with virtual methods. Calling delete on a pointer to any class derived from this one might call the wrong destructor. If any class might be a base class (by having virtual methods), then its destructor should be either be virtual or protected so that callers cannot destroy derived objects via pointers to the base.

Coding standards

Code examples

The following code example fails the check and will give a warning:

#include <iostream>

class Base
{
public:
  Base() { std::cout<< "Constructor: Base" << std::endl;}
  virtual void f(void) {}
  //non-virtual destructor:
  ~Base() { std::cout<< "Destructor : Base" << std::endl;}
};

class Derived: public Base
{
public:
  Derived() { std::cout << "Constructor: Derived" << std::endl;}
  void f(void) { std::cout << "Calling f()" << std::endl; }
  virtual ~Derived() { std::cout << "Destructor : Derived" << std::endl;}
  };   

int main(void)
{
  Base *Var = new Derived();
  delete Var; 
  return 0;
}

The following code example passes the check and will not give a warning about this issue:

#include <iostream>

class Base
{
public:
  Base() { std::cout << "Constructor: Base" << std::endl;}
  virtual void f(void) {}
  virtual ~Base() { std::cout << "Destructor : Base" << std::endl;}
};

class Derived: public Base
{
public:
  Derived() { std::cout << "Constructor: Derived" << std::endl;}
  void f(void) { std::cout << "Calling f()" << std::endl; }
  ~Derived() { std::cout << "Destructor : Derived" << std::endl;}
  };   

int main(void)
{
  Base *Var = new Derived();
  delete Var; 
  return 0;
}