英文:
Unshadow global identifier in a class
问题
我有以下的C++代码：```cppclass 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:```cppclass ident {};class Base {public:int ident;};class Derived : public Base {::ident var; // refers to the global identifierident 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）};
英文:
You might add alias to hide base member:
class Derived : public Base {using ident = ::ident;::ident var1; // OK, as beforeident 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。
英文:
If you make the base depend on a template parameter:
template <typename T>class DerivedLow : public T{void method(){ident = 0; // Refers to the global entity.}};using Derived = DerivedLow<Base>;
Then nothing from the base class will be accessible in the derived class by default. You can still access base members with this-> or Base::, or by importing them with using.
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