Template assignment operator overloading mystery

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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>&.

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