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() {
开发者_运维技巧y a;
x b = a;
b.hello(); // prints x
x* c = &a;
c->hello(); // prints y
return 0;
}
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')
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
'.
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.
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.
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