使虚拟的实现也虚拟？
在 C++ 中实现纯虚函数时，是否有最佳实践指南表明实现也应该是虚拟的？理由是什么？
class Interface
{
public:
virtual void foobar() = 0;
};
class Concrete
: public Interface
{
public:
virtual void foobar();
};
When implementing a pure virtual function in C++, is there a best-practices guideline that says the implementation should also be made virtual? What is the rationale?
class Interface
{
public:
virtual void foobar() = 0;
};
class Concrete
: public Interface
{
public:
virtual void foobar();
};
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评论（2）
不要紧。
Concrete中的void foobar()是virtual，无论您是否这样声明它，并且它会覆盖void foobar()> 在界面中。It does not matter.
void foobar()inConcreteisvirtualregardless whether you declare it as such and it overrides thevoid foobar()inInterface.尽管派生类中是否存在
virtual关键字并不重要，但我发现它是始终 包含它，以便两年后使用您的代码的任何人都可以立即看到该类的内容比立即看到的要多。Although it doesn't matter if the
virtualkeyword is present in a derived class or not, I've found it to be an indispensable time-saving self-documenting practice to always include it, so that anyone working with your code a two years from now immediately can see that there is more to the class than what immediately meets the eye.