將指針指向從私有基類別別名的成員函數

Question

以下代碼無法在Linux上使用GCC 7.2.0和Clang 5.0.0進行編譯。

#include <iostream> 

struct A 
{ 
    void f() 
    { 
     std::cout << "Hello, world!\n"; 
    } 
}; 

struct B : private A 
{ 
    using A::f; 
}; 

int main() 
{ 
    B b; 
    void (B::*f)() = &B::f; // Error: 'A' is an inaccessible base of 'B' 
    (b.*f)(); 
} 

這是否符合標準？ B中的公開使用聲明不應允許透明地使用指向B::f的成員函數指針，而不是在B的角度之外涉及A::f的可訪問性？

Answer 1

是的，你的程序不合格。

C++ 17（N4659）[namespace.udecl]/16（重點煤礦）：

For the purpose of overload resolution, the functions that are introduced by a using-declaration into a derived class are treated as though they were members of the derived class. In particular, the implicit this parameter shall be treated as if it were a pointer to the derived class rather than to the base class. This has no effect on the type of the function, and in all other respects the function remains a member of the base class.

換句話說，所述using聲明不添加的B一個構件，它只是爲同一個成員添加第二個名字A::f。這個第二個名字可以通過名稱查找來選擇，並用於名稱使用的可訪問性檢查，但除此之外，除了重載解析所指出的，它與原始成員相同。

[expr.unary.op/3：

The result of the unary & operator is a pointer to its operand. The operand shall be an lvalue or a qualified-id. If the operand is a qualified-id naming a non-static or variant member m of some class C with type T , the result has type "pointer to member of class C of type T " and is a prvalue designating C::m .

因此，即使合格-ID您使用的是拼寫的class B名稱和限定名稱查找B::f發現通過引入的名字所述using聲明在B，實際函數表達式名稱是A的成員，所以表達式&B::f具有類型「指針）返回void（類型函數的A類的成員」，或作爲type-id，void (A::*)()。您可以通過添加到您的例子驗證這一點：

#include <type_traits> 
static_assert(std::is_same<decltype(&B::f), void (A::*)()>::value, "error"); 

最後，在[conv.mem/2：

A prvalue of type "pointer to member of B of type cvT ", where B is a class type, can be converted to a prvalue of type "pointer to member of D of type cvT ", where D is a derived class of B . If B is an inaccessible, ambiguous, or virtual base class of D , or a base class of a virtual base class of D , a program that necessitates this conversion is ill-formed.

所以命名的成員函數指針是有效的，但是從void (A::*)()將其轉換到void (B::*)()不是，因爲繼承是從main無法訪問。

作爲一種變通方法，您可以提供訪問成員函數指針，除了成員函數本身B：

struct B : private A 
{ 
    using A::f; 
    using func_ptr_type = void (B::*)(); 
    static constexpr func_ptr_type f_ptr = &A::f; 
}; 

或者，如果你真的必須使用C樣式轉換。在某些情況下，允許C風格的轉換忽略類繼承關係的可訪問性，其中C++風格轉換無法用相同的結果進行編譯。 （reinterpret_cast也可以將任何指向成員函數的指針轉換爲任何其他指針，但其結果未指定。）

int main() 
{ 
    B b; 
    void (B::*f)() = (void (B::*)()) &B::f; 
    (b.*f)(); 
} 

---

在連接到這一問題的意見。注意：如果你改變main到

那麼f類型是void (A::*)()如上所述。但你碰上[expr.mptr.oper/2（重點煤礦再次）：

The binary operator .* binds its second operand, which shall be of type "pointer to member of T " to its first operand, which shall be a glvalue of class T or of a class of which T is an unambiguous and accessible base class.

所以，你還是有問題的成員函數與原來的類關聯，不能被認爲是B的成員除B和任何朋友的範圍外。