virtual assignment operator C++


Question

Assignment Operator in C++ can be made virtual. Why is it required? Can we make other operators virtual too?

1
65
3/21/2009 8:42:23 PM

The assignment operator is not required to be made virtual.

The discussion below is about operator=, but it also applies to any operator overloading that takes in the type in question, and any function that takes in the type in question.

The below discussion shows that the virtual keyword does not know about a parameter's inheritance in regards to finding a matching function signature. In the final example it shows how to properly handle assignment when dealing with inherited types.


Virtual functions don't know about parameter's inheritance:

A function's signature needs to be the same for virtual to come into play. So even though in the following example, operator= is made virtual, the call will never act as a virtual function in D, because the parameters and return value of operator= are different.

The function B::operator=(const B& right) and D::operator=(const D& right) are 100% completely different and seen as 2 distinct functions.

class B
{
public:
  virtual B& operator=(const B& right)
  {
    x = right.x;
    return *this;
  }

  int x;

};

class D : public B
{
public:
  virtual D& operator=(const D& right)
  {
    x = right.x;
    y = right.y;
    return *this;
  }
  int y;
};

Default values and having 2 overloaded operators:

You can though define a virtual function to allow you to set default values for D when it is assigned to variable of type B. This is even if your B variable is really a D stored into a reference of a B. You will not get the D::operator=(const D& right) function.

In the below case, an assignment from 2 D objects stored inside 2 B references... the D::operator=(const B& right) override is used.

//Use same B as above

class D : public B
{
public:
  virtual D& operator=(const D& right)
  {
    x = right.x;
    y = right.y;
    return *this;
  }


  virtual B& operator=(const B& right)
  {
    x = right.x;
    y = 13;//Default value
    return *this;
  }

  int y;
};


int main(int argc, char **argv) 
{
  D d1;
  B &b1 = d1;
  d1.x = 99;
  d1.y = 100;
  printf("d1.x d1.y %i %i\n", d1.x, d1.y);

  D d2;
  B &b2 = d2;
  b2 = b1;
  printf("d2.x d2.y %i %i\n", d2.x, d2.y);
  return 0;
}

Prints:

d1.x d1.y 99 100
d2.x d2.y 99 13

Which shows that D::operator=(const D& right) is never used.

Without the virtual keyword on B::operator=(const B& right) you would have the same results as above but the value of y would not be initialized. I.e. it would use the B::operator=(const B& right)


One last step to tie it all together, RTTI:

You can use RTTI to properly handle virtual functions that take in your type. Here is the last piece of the puzzle to figure out how to properly handle assignment when dealing with possibly inherited types.

virtual B& operator=(const B& right)
{
  const D *pD = dynamic_cast<const D*>(&right);
  if(pD)
  {
    x = pD->x;
    y = pD->y;
  }
  else
  {
    x = right.x;
    y = 13;//default value
  }

  return *this;
}
47
12/29/2017 2:44:20 PM

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