防止派生类从基础中隐藏非虚函数 [英] Prevent derived classes from hiding non virtual functions from base

问题描述

考虑我有A& B，以便

  class A 
 {
 public：
 void Fun 
}; 
 
 B类：public A 
 {
 .... 
}; 
  

我作为A类的设计器有什么办法可以强制执行派生类B和其他类从A派生，被阻止（获得某种错误）从隐藏非虚函数Fun（）？

解决方案

如果您希望非 - virtual 成员函数始终以某种方式访问​​，然后简单地将其包装在命名空间范围自由函数中：

 命名空间显示{
 void foo A& o）{o.foo（）; } 
} 
  

现在，B类客户可以随时进行

  void bar（）
 {
 B o; 
 revealed :: foo（o）;但是，无论B类引入了多少隐藏重载，客户端也可以只是一个简单的方法。 do 
  void bar2（）
 {
 B o; 
 A& ah = o; 
 ah.foo（）; 
} 
  
，他们可以执行
  void bar3（）
 {
 B o; 
 o.A :: foo（）; 
} 
  
所以只是一个更容易理解的符号和意图
 
 
 
也就是说，没有可能，因为评论会有它，可用性是你默认的… 
 
Consider that I have classes A & B such that
class A
{
   public:
   void Fun();
};

class B  : public A
{
   ....
};
Is there any way that I as a designer of class A can enforce that derived class B and other classes which derive from A, are prevented(get a some kind of error) from hiding the non virtual function Fun()? 
解决方案
If you want the non-virtual member function to always be accessible in some way, then simply wrap it in a namespace scope free function:
namespace revealed {
    void foo( A& o ) { o.foo(); }
}
now clients of class B can always do
void bar()
{
    B o;
    revealed::foo( o );
}
However, no matter how much class B introduces hiding overloads, clients can also just do
void bar2()
{
    B o;
    A& ah = o;
    ah.foo();
}
and they can do
void bar3()
{
    B o;
    o.A::foo();
}
so just about all that's gained is an easier-to-understand notation and intent communication.

I.e., far from being impossible, as the comments would have it, the availability is what you have by default…

                        
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