私有继承：名称查找错误 [英] Private inheritance: name lookup error

问题描述

我有以下代码示例，不编译：

I have the following code example that doesn't compile:

#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);}
    };
}

gcc打印的错误是：

The error that gcc prints is:

test.cpp：6：错误：`class my :: base1'
无法访问

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

.cpp：17：错误：在此
上下文

test.cpp:17: error: within this context

现在，我可以猜到问题是： print 的声明，编译器会看到 base1 ，并认为： base1 是 derived * this 的基类子对象，但是你不能访问它！虽然我打算 base1 应该只是一个类型名称。

Now, i can guess what the problem is: when looking at the declaration of print, the compiler sees base1 and thinks: base1 is the base-class subobject of derived* this, but you don't have access to it! While i intend that base1 should be just a type name.

如何在C ++标准中看到这个是一个正确的行为，而不是编译器中的一个错误（我确信它不是一个错误;所有编译器，我检查行为这样）？

How can i see in the C++ Standard that this is a correct behavior, and not a bug in the compiler (i am sure it's not a bug; all compilers i checked behaved so)?

我应该如何解决这个错误？所有以下修复都可以工作，但我应该选择哪一个？

How should i fix this error? All the following fixes work, but which one should i choose?

void print（ class base1 * {}

void print(class base1* pointer) {}

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

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

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

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

---

b
$ b

编辑：

---

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

推荐答案

为11.1。它建议使用:: my :: base1 *来解决这个问题：

The section you're looking for is 11.1. It suggests using ::my::base1* to work around this:

[注意：在派生类中，类名将在其声明的作用域中找到inject-class-name而不是基类的名称。 inject-class-name可能比声明的范围中的基类的名称更不可访问。 - end note]

[ 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
};

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