“注解：派生类可以恢复继承成员的访问级别，但不能使访问级别比基类中原来指定的更严格或者更宽松。”
在vs2010中经过验证，这段话是错误的。具体见以下代码：
//Base.h
#pragma once
#include <iostream>
using namespace std;
class Base
{
public:
Base(void);
~Base(void);
size_t size()const{return n;}
protected:
//private:
size_t n;
};
//Base.cpp
#include "Base.h"
Base::Base(void)
{
n = 100;
}
Base::~Base(void)
{
}
//Derived.h
#pragma once
#include "base.h"
class Derived :
private Base
{
public:
Derived(void);
~Derived(void);
using Base::size;
using Base::n;
};
//Derived.cpp
#include "Derived.h"
Derived::Derived(void)
{
}
Derived::~Derived(void)
{
}
//main.cpp
#include "Base.h"
#include "Derived.h"
#include <iostream>
using namespace std;
void main()
{
Derived XX;
Base YY;
cout<<XX.size()<<endl;
cout<<XX.n<<endl;
}
这段程序是可以正常运行没有任何错误的，但是基类Base的成员n权限是protected，在派生类中用using将其提权到了public，这就证明了原书中的那段话是错误的。
但是当我把Base类的protected 成员 n权限改成private的时候却出现了错误，因此猜测：只有子类对成员具有访问权限的时候才能改变成员的访问级别。
后来在http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2008/n2540.htm看到这样一段话：
The access rules for inheriting constructors are specified in 12.9 class.inh-ctor; otherwise all All instances of the name mentioned in a using-declaration shall be accessible. In particular, if a derived class uses a using-declaration to access a member of a base class, the member name shall be accessible. If the name is that of an overloaded member function, then all functions named shall be accessible. The base class members mentioned by a using-declaration shall be visible in the scope of at least one of the direct base classes of the class where the using-declaration is specified. ...
结果证明我的猜测是正确的。