私有繼承：名稱查找錯誤

Question

我有下面的代碼示例無法編譯：

#include <stdio.h> 

namespace my 
{ 
    class base1 
    { // line 6 
    }; 

    class base2: private base1 
    { 
    }; 

    class derived: private base2 
    { 
    public: 
     // The following function just wants to print a pointer, nothing else! 
     void print(base1* pointer) {printf("%p\n", pointer);} 
    }; 
} 

那GCC是打印錯誤：

test.cpp:6: error: `class my::base1' is inaccessible

test.cpp:17: error: within this context

現在，我能猜到是什麼問題是：在查看print的聲明時，編譯器看到base1並認爲：base1是derived* this的基類子對象，但您無權訪問它！雖然我打算base1應該只是一個類型的名字。如何在C++標準中看到這是一個正確的行爲，而不是編譯器中的錯誤（我相信它不是一個bug;我檢查過的所有編譯器都表現如此）？

我該如何解決這個錯誤？所有以下修復工作，但我應該選擇哪一個？

void print(class base1* pointer) {}

void print(::my:: base1* pointer) {}

class base1; void print(base1* pointer) {}

---

編輯：

int main() 
{ 
    my::base1 object1; 
    my::derived object3; 
    object3.print(&object1); 
} 

Answer 1

你要找的部分是11.1。它建議使用::我:: BASE1 *來解決此問題：

[ Note: In a derived class, the lookup of a base class name will find the injected-class-name instead of the name of the base class in the scope in which it was declared. The injected-class-name might be less accessible than the name of the base class in the scope in which it was declared. — end note ]

[ Example: 
class A { }; 
class B : private A { }; 
class C : public B { 
A *p; 
// error: injected-class-name A is inaccessible 
:: A * q ; 
// OK 
};