c++ - enable_if 继承的成员函数的名称查找错误

Question

我偶然发现了这个奇怪的名称查找问题，其中基类成员函数似乎根本不参与重载选择，即使它是使用 using 语句导入的。基类和派生类的成员函数都是带有 enable_if_t 的 SFINAE。

我能够使用以下代码重现我的问题:https://gcc.godbolt.org/z/ueQ-kY

#include <iostream>
#include <type_traits>

class MyTag {};

struct Base
{
    template <typename RType>
    std::enable_if_t<std::is_convertible<RType, int>::value> create(RType /*&&*/ ref)
    {
        std::cout << "Base::create(RType ref)" << std::endl;
    }
};

struct Derived : Base
{
    using Base::create;

    template <typename Tag>
    std::enable_if_t<std::is_same<Tag, MyTag>::value> create(Tag /*&&*/ tag)
    {
        std::cout << "Derived::create(Tag tag)" << std::endl;
    }
};

int main()
{
    Derived d;

    d.create(MyTag());
    d.create(0); // [x86-64 clang 7.0.0 #1] error: no matching member function for call to 'create'
}

虽然 GCC 在没有警告的情况下编译上述代码，但 clang、icc 和 MSVC 无法为 d.create(0); 的调用找到合适的重载并导致构建出错。事实上，从错误消息来看，Base::create 似乎甚至没有参与重载决议。

但是，当两个成员函数之一将其参数作为转发引用时，代码在所有主要编译器上都能正常编译。

Answer 1

Gcc 是错误的，应该拒绝你的例子。

using-declaration:
    using using-declarator-list ;

[namespace.udecl]/1

Each using-declarator in a using-declaration introduces a set of declarations into the declarative region in which the using-declaration appears. The set of declarations introduced by the using-declarator is found by performing qualified name lookup ([basic.lookup.qual], [class.member.lookup]) for the name in the using-declarator, excluding functions that are hidden as described below.

排除的功能是:

[namespace.udecl]/15

When a using-declarator brings declarations from a base class into a derived class, member functions and member function templates in the derived class override and/or hide member functions and member function templates with the same name, parameter-type-list, cv-qualification, and ref-qualifier (if any) in a base class (rather than conflicting). Such hidden or overridden declarations are excluded from the set of declarations introduced by the using-declarator.

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However, when one of the two member function takes its argument as a universal reference, the code compiles fine on all major compilers.

当其中一个函数将其参数作为(转发)引用时，此模板函数不再符合隐藏条件，因为它的parameter-type-list 不同。

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OP 已经打开了一个错误报告，检查一下:
Bug 87478 - Hidden member function falsely takes part in qualified name lookup .