問題描述

為什么 C++ 編譯器不能識別 g() 和 b 是 Superclass 的繼承成員，如以下代碼所示:

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
  }
};

如果我簡化 Subclass 并且只是從 Subclass 繼承然后它編譯.當將 g() 完全限定為 Superclass::g() 和 Superclass::b 時，它也會編譯.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:如果我在超類中公開 g() 和 b ，它仍然會失敗并出現相同的錯誤.

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

推薦答案

這可以通過使用 using 將名稱拉入當前范圍來修改:

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;
  }
};

或者通過this指針訪問來限定名稱:

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，正如您已經注意到的，通過限定全名.

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

之所以有必要這樣做，是因為 C++ 不考慮用于名稱解析的超類模板(因為它們是從屬名稱并且不考慮從屬名稱).它在您使用 Superclass 時有效，因為它不是模板(它是模板的實例化)，因此它的嵌套名稱不依賴 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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