Pack expansions in using-declarations

using-declarations should be able to cope with lists and pack expansions to allow flexible injection of names.

Motivation

With variadic templates having been introduced in C++11, many language constructs were updated to operate on pack expansions. Just as some situations require derivation from a pack of classes, some require introduction of a name from a pack of classes. Consider

template <typename... T>
struct Overloader : /* […] */ {
    // […]
};

template <typename... T>
constexpr auto make_overloader(T&&... t) {
    return Overloader<T...>{std::forward<T>(t)...};
}

int main() {
    auto o = make_overloader([] (auto const& a) {std::cout << a;},
                             [] (float f) {std::cout << std::setprecision(3) << f;});
}

The implementation of Overloader is verbose and inefficient; One has to recursively introduce overloads of operator():

template <typename T, typename... Ts>
struct Overloader : T, Overloader<Ts...> {
    using T::operator();
    using Overloader<Ts...>::operator();
    // […]
};

template <typename T> struct Overloader<T> : T {
    using T::operator();
};

While it's possible to achieve logarithmic instantiation depth, the resulting implementation is needlessly complicated and still less efficient than the following, terse syntax:

template <typename... Ts>
struct Overloader : Ts... {
    using Ts::operator()...;
    // […]
};

Design decisions and impact

We propose that a using-declaration consist of a list of names, each called a using-declarator. As the proposed change is solely an enhancement, well-formed programs and their semantics are unaffected.

Proposed wording

Modify 3.3.2 [basic.scope.pdecl] ¶4 as indicated:

The point of declaration of a using-declarationdeclarator that does not name a constructor is immediately after the using-declarationdeclarator (7.3.3).

Modify 3.4.3.1 [class.qual] ¶2 (2.2) as indicated:

— in a using-declarator of a using-declaration (7.3.3) that is a member-declaration, if the name specified after the nested-name-specifier is the same as the identifier or the simple-template-id’s template-name in the last component of the nested-name-specifier,

Modify 7.3.3 [namespace.udecl] ¶1 as indicated:

A using-declaration introduces a set of declarations into the declarative region in which the using-declaration appears.

        using-declaration:
                using using-declarator-list ;
                
        using-declarator-list:
                using-declarator ...opt
                using-declarator-list , using-declarator ...opt
        
        using-declarator:
                typenameopt nested-name-specifier unqualified-id ;
	

Each using-declarator in a using-declaration⁹⁸ introduces a set of declarations into the declarative region in which the using-declaration appears. The set of declarations introduced by the using-declarationdeclarator is found by performing qualified name lookup (3.4.3, 10.2) for the name in the using-declarationdeclarator, excluding functions that are hidden as described below. If the using-declarationdeclarator does not name a constructor, the unqualified-id is declared in the declarative region in which the using-declaration appears as a synonym for each declaration introduced by the using-declarationdeclarator. [ Note: Only the specified name is so declared; specifying an enumeration name in a using-declaration does not declare its enumerators in the using-declaration’s declarative region. — end note ] If thea using-declarationdeclarator names a constructor, it declares that the class inherits the set of constructor declarations introduced by the using-declaration from the nominated base class.

---

^{98) A using-declaration with more than one using-declarator is equivalent to a corresponding sequence of using-declarations with one using-declarator.}

Modify 7.3.3 [namespace.udecl] ¶3 as indicated:

In a using-declaration used as a member-declaration, theeach using-declarator’s nested-name-specifier shall name a base class of the class being defined. If such a using-declarationdeclarator names a constructor, theits nested-name-specifier shall name a direct base class of the class being defined. [ Example:

template <typename... bases>
struct X : bases... {
	using bases::g...;
};

X<B, D> x; // OK: B::g and D::g introduced

— end example ]

Modify 7.3.3 [namespace.udecl] ¶10 as indicated:

[Example:

namespace A {
    int i;
}

namespace A1 {
    using A::i;
    using A::i, A::i; // OK: double declaration
}
    
struct B {
    int i;
};

struct X : B {
    using B::i;
    using B::i, B::i; // error: double member declaration
};
	

— end example ]

Modify 7.3.3 [namespace.udecl] ¶19 as indicated:

If a using-declarationdeclarator uses the keyword typename and specifies a dependent name (14.6.2), the name introduced by the using-declaration is treated as a typedef-name (7.1.3).

Add a bullet (4.13) to the list in 14.5.3 [temp.variadic] ¶4:

— In a using-declaration (7.3.3); the pattern is a using-declarator.

Add a bullet (8.4.2) to the list in 14.6 [temp.res] ¶8:

— lookup for a name in the template definition found a using-declaration, but the lookup in the corresponding scope in the instantiation does not find any declarations because the using-declaration was a pack expansion and the corresponding pack is empty.

SD-6 Feature Test Macro

The feature test macro __cpp_variadic_using is recommended for this feature.

Acknowledgments

Thanks to Daveed Vandevoorde for assistance in preparing the wording.