An assignment operator shall be implemented by a non-static member function with exactly one parameter.Because a copy assignment operator operator= is implicitly declared for a class if not declared by the user ,a base class assignment operator is always hidden by the copy assignment operator of the derived class.
                                                --- 摘自 [ANSI-C++ 1998 Standard]

下面有一段代码：

这段代码的结果是：
Base's operator =
按照C++标准的说明，基类的operator =应该被hidden了，可是这里为什么还是这个结果？察看汇编代码发现，实际上Base的operator=间接的在派生类的implicitly-declared Derd::operator=中被调用。问题就来了，我明明已经在派生类中自己声明了一个operator=的啊！我翻看了一些资料，都说在你自己不声明任何复制赋值操作符的时候，编译器会为你隐式生成一个，可是这儿我已经声明了一个，为什么编译器还要声明复制赋值操作符？  
    在几位兄台的提点下，我仔细了搜索了下C++标准，查找了相关内容。我的错误大概是我没搞清楚复制赋值操作符的声明形式，A user-declared copy assignment operator X::operator= is a non-static non-template member function of class X with exactly one parameter of type X, X&, const X&, volatile X& or const volatile X&。从这一点来看，void operator =(const Base & right)根本不是Derd的复制赋值操作符，所以编译器隐式声明了一个void operator =(const Derd & right)。请大家在帮我分析下这个结论有问题不。