Overriding non-virtual methods

问题

Let\'s assume this scenario in Visual C++ 2010:

#include <iostream>
#include <conio.h>

using namespace std;

class Base
{
public:
    int b;
    void Display()
    {
        cout<<\"Base: Non-virtual display.\"<<endl;
    };
    virtual void vDisplay()
    {
        cout<<\"Base: Virtual display.\"<<endl;
    };
};

class Derived : public Base
{
public:
    int d;
    void Display()
    {
        cout<<\"Derived: Non-virtual display.\"<<endl;
    };
    virtual void vDisplay()
    {
        cout<<\"Derived: Virtual display.\"<<endl;
    };
};

int main()
{
    Base ba;
    Derived de;

    ba.Display();
    ba.vDisplay();
    de.Display();
    de.vDisplay();

    _getch();
    return 0;
};

Theoretically, the output of this little application should be:

Base: Non-virtual display.
Base: Virtual display.
Base: Non-virtual display.
Derived: Virtual display.

because the Display method of the Base class is not a virtual method so the Derived class should not be able to override it. Right?

The problem is that when I run the application, it prints this:

Base: Non-virtual display.
Base: Virtual display.
Derived: Non-virtual display.
Derived: Virtual display.

So either I didn\'t understand the concept of virtual methods or something strange happens in Visual C++.

Could someone help me with an explanation?

回答1:

Yep, you are misunderstanding a little.

The method of the same name on the derived class will hide the parent method in this case. You would imagine that if this weren't the case, trying to create a method with the same name as a base class non-virtual method should throw an error. It is allowed and it's not a problem - and if you call the method directly as you have done it will be called fine.

But, being non-virtual, C++ method lookup mechanisms that allow for polymorphism won't be used. So for example if you created an instance of your derived class but called your 'Display' method via a pointer to the base class, the base's method will be called, whereas for 'vDisplay' the derived method would be called.

For example, try adding these lines:

Base *b = &ba;
b->Display();
b->vDisplay();
b = &de;
b->Display();
b->vDisplay();

...and observe the output as expected:

Base: Non-virtual display.
Base: Virtual display.
Base: Non-virtual display.
Derived: Virtual display.

回答2:

Yes you have misunderstood a little:

Pure virtual functions:

virtual void fun1()=0 -> must be overridden in the derived class

Virtual functions:

virtual void fun2() -> can be overridden

Normal functions:

void fun3() -> don't override it

In order to achieve runtime polymorphism you need to override virtual functions in c++

回答3:

I think it might be also better to look at it in the context of static vs dynamic binding.

If the method is non-virtual (it's already by default in C++ unlike Java), then the method binds to it's caller at compile time which is impossible to know the actual object that will be pointed at runtime. So, variable type is all that matters which is the 'Base'.

来源：https://stackoverflow.com/questions/11067975/overriding-non-virtual-methods

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c++

overriding