Template assignment operator overloading mystery

Question

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

template
struct Wrapper {

  Wra         


        

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>&.

Answer 2

From the C++03 standard, §12.8/9:

A user-declared copy assignment operator X::operator= is a non-static non-template member function of class X with exactly one parameter of type X, X&, const X&, volatile X& or const volatile X&.

And §12.8/10:

If the class definition does not explicitly declare a copy assignment operator, one is declared implicitly.

The fact that your operator= is a template makes it not a copy assignment operator, so the class' implicit copy assignment operator is still generated by the compiler.