私有继承：名称查找错误

Question

我有下面的代码示例无法编译：

#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是打印错误：

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

test.cpp:17: error: within this context

现在，我能猜到是什么问题是：在查看print的声明时，编译器看到base1并认为：base1是derived* this的基类子对象，但您无权访问它！虽然我打算base1应该只是一个类型的名字。如何在C++标准中看到这是一个正确的行为，而不是编译器中的错误（我相信它不是一个bug;我检查过的所有编译器都表现如此）？

我该如何解决这个错误？所有以下修复工作，但我应该选择哪一个？

void print(class base1* pointer) {}

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

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

---

编辑：

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

Answer 1

你要找的部分是11.1。它建议使用::我:: BASE1 *来解决此问题：

[ 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 
};