使用模板访问 C++ 中超类的受保护成员 [英] accessing protected members of superclass in C++ with templates

问题描述

为什么 C++ 编译器不能识别 g() 和 b 是 Superclass 的继承成员，如以下代码所示:

Why can't a C++ compiler recognize that g() and b are inherited members of Superclass as seen in this code:

template<typename T> struct Superclass {
 protected:
  int b;
  void g() {}
};

template<typename T> struct Subclass : public Superclass<T> {
  void f() {
    g(); // compiler error: uncategorized
    b = 3; // compiler error: unrecognized
  }
};

如果我简化 Subclass 并且只是从 Subclass 继承然后它编译.当将 g() 完全限定为 Superclass::g() 和 Superclass::b 时，它也会编译.我使用的是 LLVM GCC 4.2.

If I simplify Subclass and just inherit from Subclass<int> then it compiles. It also compiles when fully qualifying g() as Superclass<T>::g() and Superclass<T>::b. I'm using LLVM GCC 4.2.

注意:如果我在超类中公开 g() 和 b ，它仍然会失败并出现相同的错误.

Note: If I make g() and b public in the superclass it still fails with same error.

推荐答案

这可以通过使用 using 将名称拉入当前范围来修改:

This can be amended by pulling the names into the current scope using using:

template<typename T> struct Subclass : public Superclass<T> {
  using Superclass<T>::b;
  using Superclass<T>::g;

  void f() {
    g();
    b = 3;
  }
};

或者通过this指针访问来限定名称:

Or by qualifying the name via the this pointer access:

template<typename T> struct Subclass : public Superclass<T> {
  void f() {
    this->g();
    this->b = 3;
  }
};

或者，正如您已经注意到的，通过限定全名.

Or, as you’ve already noticed, by qualifying the full name.

之所以有必要这样做，是因为 C++ 不考虑用于名称解析的超类模板(因为它们是从属名称并且不考虑从属名称).它在您使用 Superclass 时有效，因为它不是模板(它是模板的实例化)，因此它的嵌套名称不依赖 名字.

The reason why this is necessary is that C++ doesn’t consider superclass templates for name resolution (because then they are dependent names and dependent names are not considered). It works when you use Superclass<int> because that’s not a template (it’s an instantiation of a template) and thus its nested names are not dependent names.

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