无法显式访问名称空间范围的朋友

问题描述

我今天遇到一个问题，即ADL找不到在类内部定义的类型的静态成员函数。

I had an issue today where ADL wasn't finding a static member function for a type defined inside a class.

在下面的示例中， str（foo :: Foo :: Enum）在没有显式作用域的情况下不是通过ADL定位的， foo :: Foo :: str（foo :: Foo :: Enum）

That is, in the below example, str(foo::Foo::Enum) isn't located via ADL without explicitly scoping it, foo::Foo::str(foo::Foo::Enum)

namespace foo {

struct Foo
{
    enum Enum
    {
        FOO1,
        FOO2
    };

    static const char* str(Enum e);
};

}

foo::Foo::Enum e = foo::Foo::FOO1;
const char* s = str(e);              // ADL doesn't work

我发现 SO问题，并按照接受的答案所述，将其更改为 friend 函数导致ADL现在可以工作。

I found this SO question, and as stated in the accepted answer, changing it to a friend function results in ADL now working.

namespace foo {

struct Foo
{
    enum Enum
    {
        FOO1,
        FOO2
    };

    friend const char* str(Enum e);  // note str is now a friend
};

}

foo::Foo::Enum e = foo::Foo::FOO1;
const char* s = str(e);              // ADL works now

虽然这现在对ADL有所帮助，但我惊讶地发现自己无法通过使用命名空间 foo

Whilst this now helps ADL, I was surprised to find that I couldn't access str by scoping it with a namespace foo

foo::Foo::Enum e = foo::Foo::FOO1;
const char* s = foo::str(e);         // error: ‘str’ is not a member of ‘foo’

我运行了一个测试，其中我打印了 __ PRETTY_FUNCTION __ 的结果，更惊讶地发现str的范围显然是 foo :: ：

I ran a test, where I printed out the result of __PRETTY_FUNCTION__, and was even more surprised to see that the scope of str is apparently foo:::

__PRETTY_FUNCTION__: const char* foo::str(foo::Foo::Enum)

以下工作示例：

#include <iostream>

namespace foo {

struct Foo
{
    enum Enum
    {
        FOO1,
        FOO2
    };

    friend const char* str(Enum e)
    {
        return __PRETTY_FUNCTION__;
    }
};

}

int main()
{
    foo::Foo::Enum e = foo::Foo::FOO1;

    std::cout << str(e) << '\n';
    // std::cout << foo::str(e) << '\n'; // error: ‘str’ is not a member of ‘foo’

    return 0;
}

输出：

$ ./a.out
const char* foo::str(foo::Foo::Enum)

问题：

• 为什么我无法找到 str（..）明确地用封闭的名称空间对其进行范围界定？


• 为什么 __ PRETTY_FUNCTION __ 说它在 foo :: 中，但是我却找不到它吗？

• Why am I unable to locate str(..) explicitly scoping it with the enclosing namespace?
• Why does __PRETTY_FUNCTION__ say it's in foo::, and yet I am unable to locate it as such?

推荐答案

根据标准

这意味着 str 不可见命名查找；

That means str is not visible to name lookup; it can only be called via ADL.

来自，

str 确实成为命名空间 foo 的成员；

str does become member of namespace foo; it's just invisible.

：

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