问题

我有以下的C++代码：
```cpp
class ident {};

class Base {
public:
  int ident;
};

class Derived : public Base {
  ::ident var; // 引用全局标识符
  ident var; // 语法错误，引用基类成员
};

是否有可能在Derived::method()中使ident引用全局标识符？最终完全隐藏Base::ident？我无法更改Base，但我可以创建一个中间类。我可以在任何地方添加代码行，但成员变量的生成始终将ident表示为ident。

这是一个非常不寻常的请求，因为它是用于一个模板化的代码生成器（SWIG）。

<details>
<summary>英文:</summary>

I have the following C++ code:
```cpp
class ident {};

class Base {
public:
  int ident;
};

class Derived : public Base {
  ::ident var; // refers to the global identifier
  ident var; // syntax error, refers to the base member
};

Is it possible to make it so ident refers to the global identifier in Derived::method()? Eventually completely hiding Base::ident? I cannot change Base, but I can make an intermediate class. I can add lines everywhere but the generation of the member variables will always refer to ident as ident.

It is a very unusual request, because it is for a templated code generator (SWIG).

答案1

得分: 3

你可以添加别名来隐藏基类成员：

class Derived : public Base {
  using ident = ::ident;
  ::ident var1; // OK，与之前相同
  ident var2;   // OK，引用typename Derived::ident（因此也是::ident）
};

class Derived : public Base {
  using ident = ::ident;
  ::ident var1; // OK, as before
  ident var2;   // OK, refer to typename Derived::ident (and so ::ident)
};

答案2

得分: 0

如果您将基类依赖于模板参数：

template <typename T>
class DerivedLow : public T
{
    void method()
    {
        ident = 0; // 引用全局实体。
    }
};

using Derived = DerivedLow<Base>;

那么默认情况下，派生类中将无法访问基类的任何成员。您仍然可以使用 this-> 或 Base:: 来访问基类成员，或者通过 using 导入它们。

template <typename T>
class DerivedLow : public T
{
    void method()
    {
        ident = 0; // Refers to the global entity.
    }
};

using Derived = DerivedLow<Base>;