C ++继承和功能覆盖

Question

In C++, will a member function of a base class be overridden by its derived class function of the same name, even if its prototype (parameters' count, type and constness) is different ? I guess this a silly question, since many websites says that the function prototype should be the same for that to happen; but why doesn't the below code compile? It's a very simple case of inheritance, I believe.

#include <iostream>
using std::cout;
using std::endl;

class A {};
class B {};

class X
{
public:
    void spray(A&)
    {
        cout << "Class A" << endl;
    }
};

class Y : public X
{
public:
    void spray(B&)
    {
        cout << "Class B" << endl;
    }
};

int main()
{
    A a;
    B b;
    Y y;

    y.spray(a);
    y.spray(b);

    return 0;
}

GCC throws

error: no matching function for call to `Y::spray(A&)'
note: candidates are: void Y::spray(B&)

Answer 1

The term used to describe this is "hiding", rather than "overriding". A member of a derived class will, by default, make any members of base classes with the same name inaccessible, whether or not they have the same signature. If you want to access the base class members, you can pull them into the derived class with a using declaration. In this case, add the following to class Y :

using X::spray;

Answer 2

That's so called 'hiding': Y::spray hides X::spray . Add using directive:

class Y : public X
{
public:
   using X::spray;
   // ...
};

Answer 3

Classes are scopes and a class scope is nested in its parent. You have exactly the same behavior with other nested scopes (namespaces, blocks).

What happen is that when the name lookup searches for the definition of a name, it looks in the current namespace, then in the englobing namespace and so on until it find one definition; the search then stop (that's without taking into account the complications introduced by argument dependent name lookup -- the part of the rules which allows to use a function defined in the namespace of one of its argument).