Question

I have a simple struct Wrapper, distinguished by two templated assignment operator overloads:

template<typename T>
struct Wrapper {

  Wrapper() {}

  template <typename U>
  Wrapper &operator=(const Wrapper<U> &rhs) {
    cout << "1" << endl;
    return *this;
  }
  template <typename U>
  Wrapper &operator=(Wrapper<U> &rhs) {
    cout << "2" << endl;
    return *this;
  }
};

I then declare a and b:

Wrapper<float> a, b;
a = b;

assigning b to a will use the non-const templated assignment operator overload from above, and the number “2” is displayed.

What puzzles me is this: If I declare c and d,

Wrapper<float> c;
const Wrapper<float> d;
c = d;

and assign d to c, neither of the two assignment operator overloads is used, and no output is displayed; so the default copy assignment operator is invoked. Why does assigning d to c not use the const overloaded assignment operator provided? Or instead, why does assigning b to a not use the default copy assignment operator?

Answer 1

Why does assigning d to c not use the const overloaded assignment operator provided?

The implicitly-declared copy assignment operator, which is declared as follows, is still generated:

Wrapper& operator=(const Wrapper&);

An operator template does not suppress generation of the implicitly-declared copy assignment operator. Since the argument (a const-qualified Wrapper) is an exact match for the parameter of this operator (const Wrapper&), it is selected during overload resolution.

The operator template is not selected and there is no ambiguity because–all other things being equal–a nontemplate is a better match during overload resolution than a template.

Why does assigning b to a not use the default copy assignment operator?

The argument (a non-const-qualified Wrapper) is a better match for the operator template that takes a Wrapper<U>& than for the implicitly-declared copy assignment operator (which takes a const Wrapper<U>&.