Example code:
public class A {
public int number;
public A(int number) {
this.number = number;
}
public int getNumber() {
return number;
}
}
public class B extends A{
public int number;
public B(int number) {
super(number);
}
public int getNumber() {
return number;
}
}
public class C {
public static void main(String args[]) {
A test1 = new B(2);
B test2 = new B(2);
System.out.println(test1.number) // prints 2
System.out.println(test2.number) // prints 0
System.out.println(test1.getNumber()) //prints 0
System.out.println(test2.getNumber()) // prints 0
}
}
As shown above test1.number is not equal to test1.getNumber().
So when I make test1 an object of type A test1.number is referring to the int number in class A.
But when I call test1.getNumber() it's calling getNumber() in class B?
Why does that happen?
All methods in Java are virtual. I'll get to what that means in a second.
So, on this line:
A test1 = new B(2);
You've assigned a new instance of class B to a variable declared as holding an A.
When you use test1.number, you're using the number variable declared in class A, which given that the constructor for B calls super(number) will be 2.
However, when you call test1.getNumber(), that's where the virtual kicks in. Virtual means that it will always call a method in the constructed class rather than the type of variable that you've declared it as. In other words, rather than calling A's getNumber like you thought it would, it actually call's B's getNumber instead. B's constructor doesn't assign a value to B's number variable, so you get 0.
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