C++ pure virtual function have body [duplicate]

Answer 1

Your assumption that pure virtual function cannot be called is absolutely incorrect. When a function is declared pure virtual, it simply means that this function cannot get called dynamically, through a virtual dispatch mechanism. Yet, this very same function can easily be called statically, non-virtually, directly (without virtual dispatch).

In C++ language a non-virtual call to a virtual function is performed when a qualified name of the function is used in the call, i.e. when the function name specified in the call has the <class name>::<function name> form.

For example

struct S 
{
  virtual void foo() = 0;
};

void S::foo() 
{
  // body for pure virtual function `S::foo`
}

struct D : S 
{
  void foo() 
  {
    S::foo();       
    // Non-virtual call to `S::foo` from derived class

    this->S::foo(); 
    // Alternative syntax to perform the same non-virtual call 
    // to `S::foo` from derived class
  }
};

int main() 
{
  D d;

  d.S::foo(); 
  // Another non-virtual call to `S::foo`
}

Answer 2

For most pure virtual functions, you'd be right. However, for a pure virtual destructor, it's actually important to define a corresponding destructor implementation:

• The "pure virtual" is to require derived classes to implement their destructor.
• Your base class destructor implementation is so that the derived class destructors can successfully "chain up" afterwards.

Answer 3

"Effective C++" Meyers mentions a reason for a pure virtual function to have a body: Derived classes that implement this pure virtual function may call this implementation smwhere in their code. If part of the code of two different derived classes is similar then it makes sense to move it up in the hierarchy, even if the function should be pure virtual.

see here.