C++ assignment operator in derived class

Question

I don't quite understand why for an assignment a derived class do not call the corresponding operator of the base class, if its own does not exist. Look at the code:

#include <iostream>
using namespace std;

class A{
protected:
  void myWrite(){
    cout << " Base " <<endl;
  }
public:
  double x,y;
  A(): x{0},y{0}{};
  virtual A & operator=(double g){x=g;y=g;return *this;}
  virtual ~A(){};
  virtual void doneit(){myWrite();}
};


class B: public A{
protected:
public:
  B():A(){};
  virtual ~B(){};
  virtual void doneit(){myWrite();}
};

int main() {
  A jj;
  B pp;

  pp=0.0;
  return 0;
}

As it is the code does not compile. Of course if I define an "operator=" for B identical to that for A, everything works, but why the B "operator=" is not called by default if the one in the derived class is not defined?? Could you help shed some light on this issue?

The gcc compiler says ../src/tito.cpp:40:4: error: no viable overloaded '=' pp=0.0; ~~^~~~ ../src/tito.cpp:28:7: note: candidate function (the implicit copy assignment operator) not viable: no known conversion from 'double' to 'const B' for 1st argument class B: public A{ ^ 1 error generated.

Could you explain why it doesn't work?

score 24 · Answer 1 · answered Jun 24 '15 at 14:24

24

If you don't declare a copy assignment operator, the compiler will declare one for you. So your class B really looks like:

class B : public A {
public:
    // explicit
    B();
    virtual ~B();
    virtual void doneit();

    // implicit
    B(const B&);
    B& operator=(const B&);
};

The implicit copy assignment operator hides the A::operator=(double) that you wrote, so it's the only candidate that name lookup will find. Of course, it's not a viable candidate, since double isn't convertible to B, hence the error.

To use the A::operator=(double ) operator, you have to explicitly bring it into scope:

using A::operator=;

But then you're not going to be assigning any of the B part. So it'd be better to be more explicit:

B& operator=(double g) {
    // B stuff here

    // A stuff
    A::operator=(g);

    return *this;
}

answered Jun 24 '15 at 14:24

Barry

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1

Yes, the main rationale is that parent's `operator =` does not assign the child's part so implicitly using it can easilly create broken objects. – StenSoft Jun 24 '15 at 14:26
Thank you for the very useful comment!! – Max Jun 25 '15 at 15:06
1

I would add the assignment of the B stuff after the A stuff, so that the self-check, which is done when implementing B& operator=(double A), is done at the beginning for both `operator=` functions (as the self-check should be done only at the base class). – jaques-sam Nov 05 '18 at 13:52
@DrumM What self-check? We're assigning from a `double`. – Barry Nov 05 '18 at 15:13

Vlad from Moscow · Answer 2 · 2015-06-24T14:52:36.137

The copy assignment operator of the derived class that is implicitly declared by the compiler hides assignment operators of the base class. Use using declaration in the derived class the following way

class B: public A{
protected:
public:
  using A::operator =;     
  B():A(){};
  virtual ~B(){};
  virtual void doneit(){myWrite();}
};

Another approach is to redeclare the virtual assignment operator in the derived class

B & operator=(double g) { A::operator =( g ) ;return *this;}

In this case you can use the polymorphism. For example

#include <iostream>

class A
{
public:
  double x,y;
  A(): x{0}, y{0} {}
  virtual A & operator =( double g ){ x = g; y = g; return *this; }
  virtual ~A(){}
};


class B: public A
{
public:
  B() : A(){}
  virtual B & operator =( double g ) override { A::operator =( 2 * g ) ; return *this; }
  virtual ~B(){}
};

int main() 
{
    A a;
    B b;

    A *p = &a;

    *p = 10;

    std::cout << p->x << ' ' << p->y << std::endl; 

    p = &b;

    *p = 10;

    std::cout << p->x << ' ' << p->y << std::endl; 

    return 0;
}

The program output is

10 10
20 20

C++ assignment operator in derived class

2 Answers2