将嵌套类导入命名空间 - C ++

问题描述

说我有这样的类：

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

我可以参考 B c $ c> A :: B 和 f（） as A :: f $ c>，但是我可以导入 B 和 f（）到全局/当前命名空间？我尝试了

 使用A :: B; 
  

但是给我一个编译错误。

解决方案

您应该可以为类使用命名空间别名：

  = A :: B; 
  

但是你不能使用成员函数，甚至不能使用静态成员函数。 p>



修改：根据这个应该是有效的，实际上创建一个类型别名的方式与 typedef 。

---

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

  struct Foo 
 {
 static void bar b $ b {} 
}; 
 
 auto bar = Foo :: bar; 
  

编辑：当然，有一个全局变量指向一个静态成员函数在旧的C ++标准中也是可能的，但它比使用C ++ 11的 auto 关键字更麻烦。在上面的例子中，它将是：

  void（* bar）（）= Foo :: bar; 
  

Say I have a class like this:

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

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;

but that gave me a compilation error.

解决方案

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

using B = A::B;

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

Edit: 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.

---

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;

Edit: 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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