Reputation: 20312
Question with virtual functions
I have two classes:
class x {
public:
virtual void hello() {
std::cout << "x" << std::endl;
}
};
class y : public x {
public:
void hello() {
std::cout << "y" << std::endl;
}
};
Can someone explain why the following two calls to hello() print different messages? Why don't they both print "y"? Is it because the first one is a copy while the second one actually points to the object in memory?
int main() {
y a;
x b = a;
b.hello(); // prints x
x* c = &a;
c->hello(); // prints y
return 0;
}
Upvotes: 7
Views: 296
Answers (4)
Reputation: 5696
b.hello() prints "x" because b is an instance of class X. c->hello() prints "y" because c points to a, and a is an instance of class Y.
What might be confusing for you is that when you write x b = a;, you're creating a new object b and initializing it with a. When you write x* c = &a;, c is not a new object. You just introduced an alias to an existing object.
Upvotes: 0
Reputation: 101456
Because x b = a; slices the object.
When this code executes, it creates a new x, not a y, which is a copy of the original y, a'.
Upvotes: 6
Reputation: 67193
x b = a copies a to b. Since b is type x, you end up with an object of type x. An object of type x will print x.
The only way you get y is when you are calling into an object of type y.
Upvotes: 0
Reputation: 107247
Yes, you are right
x b = a;
Invokes a copy constructor (b IS an 'x')
x& b = a;
Assigns a reference and will use the override (b is still actually a 'y')
Upvotes: 6