I have a class Child
that extends Parent
.
Parent child = new Child();
if (child instanceof Parent){
// Do something
}
<
Yes, it would. And why should it not?
Because child is in fact an instance of Parent. If, you want to perform an operation only for a child you should check
if (child instanceof Child){
}
However you should remember the following statement from Effective C++, by Scott Meyers :
"Anytime you find yourself writing code of the form "if the object is of type T1, then do something, but if it's of type T2, then do something else," slap yourself.
which I think applies in this case too. If you want to doSomething based on what type of class the referenced object belongs to, the following code structure should help you with it.
NOTE: I have not compiled it.
class Parent {
public void doSomething() {
System.out.println("I am the Parent, and I do as I like");
}
}
class ChildA extends Parent {
public void doSomething() {
System.out.println("I am a child named A, but I have my own ways, different from Parent");
}
}
class ChildB extends Parent {
public void doSomething() {
System.out.println("I am a child named B, but I have my own ways, different from my Parent and my siblings");
}
}
public class Polymorphism101 {
public static void main(String[] args) {
Parent p = new Parent();
p.doSomething();
p = new ChildA();
p.doSomething();
p = new ChildB();
p.doSomething();
}
}
EDIT: A better example
You could be developing a drawing application. An application that draws shapes of any kind. In that case, you should have an abstract type Shape
.
For purpose(s) like; drawing all shapes; list all shapes; find a shape or delete a shape, you need to have a list of Shapes. Since the list is of a parent type, it can store any shapes.
The Shape
interface/abstract class/virtual class should have an abstract/pure virtual function Draw()
. So, in your DrawToDeviceLoop, you just call Draw()
for each shape, you never need to check what shape it is.
The Shape
interface can have an abstract implementation AbstractShape
, which can have shape name or id as data members and GetName, Cleanup and other functions with functionality common to all shapes.
Remember an abstract type cannot be instantiated, so Shape
itself cannot be instantiated, as it cannot be drawn either.
ofcourse it returns true because child is an instance of the parent
instanceof will return true if it's a subclass...
instanceof Documentation
Yes. instanceof
will be true whenever the reference(left side of the instanceof
expression) can be cast to the ReferenceType(the type on the right side of the instanceof
expression). This will be true for subclasses in relation to their parent:
Child child = new Child();
Parent parent = (Parent) child; //works!
assert child instanceof Parent; //true
From The Java Language Specification, Java SE 9 Edition:
15.20. Relational Operators
...
RelationalExpression instanceof ReferenceType15.20.2. Type Comparison Operator instanceof
...
At run time, the result of theinstanceof
operator istrue
if the value of the RelationalExpression is notnull
and the reference could be cast to the ReferenceType without raising aClassCastException
. Otherwise the result isfalse
.