Virtual function keyword

问题:

Is there any difference between declaring inherited virtual function in a child class with the "virtual" keyword or not, considering I want to call fun appropriate to my objects' type. Look at the comments.

#include <cstdio>
struct A{
    int a;
    A():a(5){}
    virtual int fun(){return a+1;}
};
struct B: public A{
    virtual int fun(){return a+5;} //I put virtual here
//  int fun(){return a+5;} // Any difference if I put virtual before or not?
};
int main(){
    B obj;
    printf("%d\n", static_cast<A>(obj).fun()); // A::fun() called. Why?
    printf("%d\n", static_cast<A&>(obj).fun()); // B::fun() called. As expected
    printf("%d\n", static_cast<A*>(&obj)->fun()); // B::fun() called. As expected
    printf("%d\n", static_cast<A>(B()).fun()); // A::fun() again. Why?
//  printf("%d\n", static_cast<A&>(B()).fun()); //invalid_cast error. Why? 
    printf("%d\n", static_cast<A*>(&B())->fun()); //It works! B::fun() call
    return 0;
}

回答1:

Overriding functions in derived classes are implicitly declared "virtual" if the corresponding function in the base class is virtual. Just make sure you got the exact same signature, or you might inadvertently hide the original function and declare a new one!

In C++0x, feel free to make liberal use of the override specifier.

Your two "Why?" questions are because of slicing; you're making new, copy-sliced objects of type A. Note that in B x; static_cast<A>(x); the cast is the same as saying A(x).

回答2:

Keeping the virtual key word before the overridden member function in the derived class is optional. Run-time polymorphism works only for pointers or references.