使用模板访问C ++中的超类的受保护成员 [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 $ c>如下代码中所示的超类：

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
  }
};

如果我简化子类 子类< int> 然后它编译。当完全限定 g（）为超类 :: g（）和超类< T> :: b 。我使用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：如果我使 g（）和

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

推荐答案

这可以通过修改使用使用将名称拉入当前范围：

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;
  }
};

或通过 this 指针访问：

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

或者，您已经注意到，通过限定全名。

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

这是必要的原因是C ++不考虑名称解析的超类模板（因为它们是依赖名称，不考虑依赖名称）。当您使用超类时，因为它不是模板（它是模板的一个实例化），因此它的嵌套名称不是依赖名称。

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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