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Answer:
Answer
A virtual method is a method that you should override when inheriting that class
It uses virtual as a keyword.
Answer
Virtual functions allow polymorphism. If a function is not declared virtual it will be impossible to override it in a child class. Example:
class Foobar
{
public:
int i;
Foobar() {i = -1;}
virtual void FuncA() {i = 0;}
void FuncB() {i = 10;}
Foobar() {i = -1;}
virtual void FuncA() {i = 0;}
void FuncB() {i = 10;}
};
class Foo: public Foobar
{
public:
virtual void FuncA() {i = 1;}
void FuncB() {i = 11;}
void FuncB() {i = 11;}
};
class Bar: public Foobar
{
public:
virtual void FuncA() {i = 2;}
void FuncB() {i = 12;}
void FuncB() {i = 12;}
};
void FoobarFunc(Foobar* fb)
{
cout << fb->i << endl;
fb->FuncA();
cout << fb->i << endl;
fb->FuncB();
cout << fb->i << endl << endl;
fb->FuncA();
cout << fb->i << endl;
fb->FuncB();
cout << fb->i << endl << endl;
}
void FooFunc(Foo* f)
{
cout << f->i << endl;
f->FuncA();
cout << f->i << endl;
f->FuncB();
cout << f->i << endl << endl;
f->FuncA();
cout << f->i << endl;
f->FuncB();
cout << f->i << endl << endl;
}
When FoobarFunc is called with Foo, Bar, or Foobar, the line fb->FuncA() will call the the correct FuncA from each class according to polymorphism. However, in FoobarFunc, the line fb->FuncB() will ALWAYS call Foobar's FuncB no matter what type is actually passed in. To further illustrate this, in FooFunc, the line f->FuncB() will ALWAYS call Foo's FuncB.
First answer by ComputerJy.
Last edit by Sudennis315.
Contributor trust: 30
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Question popularity:
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