将嵌套类导入命名空间 - C++

2022-01-14 00:00:00 namespaces c++

假设我有这样的课程:

class A {
public:
    class B {
        // ...
    };
    static void f();
    // ...
};

我可以将 B 称为 A::B 并将 f() 称为 A::f()，但我可以将 B 和 f() 导入全局/当前命名空间吗?我试过了

I can refer to B as A::B and to f() as A::f(), but can I import B and f() into the global/current namespace? I tried

using A::B;

但这给了我一个编译错误.

but that gave me a compilation error.

推荐答案

你应该能够为类使用命名空间别名:

You should be able to use namespace aliases for the class:

using B = A::B;

但是你不能用成员函数做到这一点，甚至不能用静态成员函数.

However you can't do that with the member function, not even with static member functions.

根据this SO answer(What is the difference between 'typedef' and 'using'在 C++11) 这个 应该 是有效的，并且实际上以与 typedef 相同的方式创建类型别名.但是，它只是 C++11.

According to this SO answer (What is the difference between 'typedef' and 'using' in C++11) this should be valid, and actually creates a type alias in the same way that typedef does. However, it's C++11 only.

在 C++11 中有一个静态成员函数的解决方法，通过声明一个指向静态函数的变量:

There is a workaround for static member functions in C++11, by declaring a variable pointing to the static function:

struct Foo
{
    static void bar()
        { }
};

auto bar = Foo::bar;

当然，在旧的 C++ 标准中也可以有一个指向静态成员函数的全局变量，但它比使用 auto 更混乱C++11 的关键字.在上面的示例中，它将是:

Of course, having a global variable pointing to a static member function is possible in the older C++ standard as well, but it's more messy than using the auto keyword of C++11. In the example above it would be:

void (*bar)() = Foo::bar;

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