ang bug？命名空间模板类“的朋友 [英] clang bug? namespaced template class' friend

问题描述

下面的代码不是在clang下编译，而是在gcc和VS下编译：

The following code which doesn't compile under clang but does under gcc and VS:

template<typename T> class bar;

namespace NS
{
    template<typename T>
    class foo
    {
        foo() {}

        template<typename U> friend class bar;
    };
}

template<typename R>
class bar
{
public:
    bar()
    {
        NS::foo<int> f;
    }
};


int main(int, char **)
{
    bar<int> b;        
    return 0;
}

无效：

main.cpp:20:22: error: calling a private constructor of class 'NS::foo<int>'

        NS::foo<int> f;    
                     ^

main.cpp:8:9: note: implicitly declared private here

        foo() {}   
        ^

bar 应该可以访问 foo 的私有构造函数，但它看起来没有。如果我删除命名空间NS ，它会编译。

bar should have access to foo's private constructor but it looks like it doesn't. If I remove namespace NS, it compiles.

代码看起来很好，但也许我误解了C ++标准。哪个编译器是正确的？

Code looks fine to me, but maybe I'm misunderstanding the C++ standard. Which compiler is correct?

推荐答案

我相信clang是正确的。根据[namespace.memdef] / 3：

I believe that clang is correct. According to [namespace.memdef]/3:

命名空间中首先声明的每个名称都是该命名空间的成员。如果
非本地类中的 friend 声明首先声明类，函数，类模板或函数模板，则该朋友是成员
最内层的命名空间。

Every name first declared in a namespace is a member of that namespace. If a friend declaration in a non-local class first declares a class, function, class template or function template the friend is a member of the innermost enclosing namespace.

在您的情况下，名称不会显示为由 friend 声明。然而，在该段落的后面，强调我：

In your case, the name wouldn't appear to be "first declared" by the friend declaration. Later in that paragraph, however, emphasis mine:

如果朋友声明既不合格也不是一个 template-id 和
声明是一个函数或一个 elaborated-type-specifier ，查找来确定实体是否
之前已声明不会考虑最内层命名空间之外的任何作用域。

If the name in a friend declaration is neither qualified nor a template-id and the declaration is a function or an elaborated-type-specifier, the lookup to determine whether the entity has been previously declared shall not consider any scopes outside the innermost enclosing namespace.

，该声明：

template<typename U> friend class bar;

不会在 bar code> namespace NS ，所以它不会找到你的早期声明。因此，它声明一个类模板 NS :: bar< typename> 是朋友 $ c> foo 。您必须限定 bar 的名称才能找到

will not look for bar outside of namespace NS, so it will not find your earlier declaration. As such, it declares a class template NS::bar<typename > to be a friend of foo. You will have to qualify the name bar in order for it to be found:

template<typename U> friend class ::bar;

这似乎与 GCC Bug 37804 。

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