奇怪的編譯器錯誤和模板繼承

Question

有人能解釋我爲什麼這個代碼：

class safe_bool_base 
{ //13 
    protected: 

     typedef void (safe_bool_base::*bool_type)() const; 

     void this_type_does_not_support_comparisons() const {} //18 

     safe_bool_base() {} 
     safe_bool_base(const safe_bool_base&) {} 
     safe_bool_base& operator=(const safe_bool_base&) { return *this; } 
     ~safe_bool_base() {} 
}; 

template <typename T=void> class safe_bool : public safe_bool_base 
{ 
    public: 

     operator bool_type() const 
     { 
      return (static_cast<const T*>(this))->boolean_test() ? &safe_bool_base::this_type_does_not_support_comparisons : 0; 
     } 

    protected: 

     ~safe_bool() {} 
}; 

template <> class safe_bool<void> : public safe_bool_base 
{ 
    public: 

     operator bool_type() const 
     { 
      return (boolean_test() == true) ? &safe_bool_base::this_type_does_not_support_comparisons : 0; //46 
     } 

    protected: 

     virtual bool boolean_test() const = 0; 
     virtual ~safe_bool() {} 
}; 

產生如下的編譯器錯誤？

c:\project\include\safe_bool.hpp(46) : error C2248: 'safe_bool_base::this_type_does_not_support_comparisons' : cannot access protected member declared in class 'safe_bool_base' 
c:\project\include\safe_bool.hpp(18) : see declaration of 'safe_bool_base::this_type_does_not_support_comparisons' 
c:\project\include\safe_bool.hpp(13) : see declaration of 'safe_bool_base' 

由於兩個safe_bool模板從safe_bool_base派生，我不明白爲什麼人們不能訪問基類的保護成員。

我錯過了什麼嗎？

Answer 1

這也許應該幫助（在非模板的情況也可重複）

struct A{ 
protected: 
    void f(){} 
}; 

struct B : A{ 
    void g(){&A::f;}  // error, due to Standard rule quoted below 
}; 

int main(){ 
} 

VS gives "'A::f' : cannot access protected member declared in class 'A'"

對於相同的代碼，科莫給

"ComeauTest.c", line 7: error: protected function "A::f" (declared at line 3) is not accessible through a "A" pointer or object void g(){&A::f;} ^

"ComeauTest.c", line 7: warning: expression has no effect void g(){&A::f;}

這是固定的代碼，它實現了所需的意圖

struct A{ 
protected: 
    void f(){} 
}; 

struct B : A{ 
    void g(){&B::f;}  // works now 
}; 

int main(){ 
} 

那麼，爲什麼第一個代碼片段不起作用？

這是因爲在C++ Standard03

11.5/1- "When a friend or a member function of a derived class references a protected nonstatic member function or protected nonstatic data member of a base class, an access check applies in addition to those described earlier in clause 11.102) Except when forming a pointer to member (5.3.1), the access must be through a pointer to, reference to, or object of the derived class itself (or any class derived from that class) (5.2.5). If the access is to form a pointer to member, the nested-name-specifier shall name the derived class (or any class derived from that class).

以下規則的改變如此操作功能內的回報如下

return (boolean_test() == true) ? &safe_bool<void>::this_type_does_not_support_comparisons : 0; //46 

return (static_cast<const T*>(this))->boolean_test() ? &typename safe_bool<T>::this_type_does_not_support_comparisons : 0; 

編輯2：請忽略我的解釋。大衛是對的。這是它歸結爲什麼。

struct A{ 
protected: 
    int x; 
}; 

struct B : A{ 
    void f(); 
}; 

struct C : B{}; 

struct D: A{   // not from 'C' 
}; 

void B::f(){ 
    x = 2;   // it's own 'A' subobjects 'x'. Well-formed 

    B b; 
    b.x = 2;  // access through B, well-formed 

    C c; 
    c.x = 2;  // access in 'B' using 'C' which is derived from 'B', well-formed. 

    D d; 
    d.x = 2;  // ill-formed. 'B' and 'D' unrelated even though 'A' is a common base 
} 

int main(){} 

Answer 2

我不認爲這與模板有關。您的示例代碼可以減少此，它仍然給出了相當於錯誤：

class A 
{ 
    protected: 
     typedef void (A::*type)() const; 
     void foo() const {} 
}; 


class B : public A 
{ 
    public: 
     operator type() const 
     { 
      return &A::foo; 
     } 
}; 

我相信這個問題是你不能在公共接口成員函數指針返回protected成員。（編輯：不是真的...）

Answer 3

Chubsdad的回答澄清了爲什麼模板專業化存在錯誤的問題。

現在下面的C++標準規則

14.7.2/11 通常的訪問檢查規則並不適用於用來指定明確的
實例名。 [注意：特別是，函數
聲明符（包括參數類型，返回類型和異常規格）中使用的模板參數和名稱可能是
通常不可訪問的私有類型或對象，模板可能是
成員模板或成員函數通常無法訪問。 - 尾註]

將解釋爲什麼泛型模板實例化不會拋出錯誤。即使你有私人訪問說明符也不會拋出。