Inheritance and Casting

[Thomas Hauck]

The following diagram is given below along with the corresponding assignments:



In the above example
1) line 1 is given
2) line 2 represents Upcasting (You are assigning a subclass reference "x" to a parent class reference "a")
3) line 3 causes a ClassCastException because variable "x" is not a Human
4) Line 4 compiles and runs (reference variable x is referred to by an "Animal" reference variable "a")
5) Line 5 compiles and runs
6) line 6 causes a ClassCastException because Mammal is not a Human
7) line 7- Type mismatch: cannot convert from Animal to Mammal

interface Roamable { }
class Animal implements Roamable { }
class Mammal extends Animal {}

public class Human extends Mammal {
	public static void main(String[] args) {
		Mammal x = new Mammal();                // line 1 
		Animal a = x;                            // line 2
		//Roamable f = (Human) x;				// line 3 - ClassCastException
		Roamable f1 = (Animal) x; 				// line 4 - Compiles and Runs
		Mammal b = (Mammal) (Animal)x;          // line 5 - Compiles and Runs
		//Roamable f2 = (Human) new Mammal(); 	// line 6 - ClassCastException
		//Mammal b1 = (Animal)x; 				// line 7-  Compiler Error
	}
}

/* 
 line 3 - causes  
 Exception in thread "main" java.lang.ClassCastException: com.java.inheritance.Mammal cannot be cast to com.java.inheritance.Human
at com.java.inheritance.Human.main(Human.java:11)
-----
line 6 - causes	
Exception in thread "main" java.lang.ClassCastException: com.java.inheritance.Mammal cannot be cast to com.java.inheritance.Human 
at com.java.inheritance.Human.main(Human.java:14)
-----
line 7- Type mismatch: cannot convert from Animal to Mammal
*/

Question: How does the double cast on line 5 work?

Mammal b = (Mammal) (Animal)x; // line 5 - Compiles and Runs

Lines 5 and 7 are almost the same except for line 5 contains (Mammal).
I am interested in knowing what enables line 5 to compile and run.

[Roel De Nijs]

Line 5

Mammal b = (Mammal) (Animal)x;

can be rewritten to:

Animal temp = (Animal)x;
Mammal b = (Mammal)temp;

Thomas Hauck wrote:I am interested in knowing what enables line 5 to compile and run.

Why do you think it would not compile? x is a Mammal. So first of all you cast it explicitly to an Animal (that's similar to line 2, where the cast happens implicitly). And then you cast the temporary result back to a Mammal (the object to which the temporary result is referring is still a Mammal, so no problem at all).

If in line 7 the type of variable b1 would be changed to Animal (instead of Mammal) that line would compile and run without any problem (and you have exactly the same line as in the rewritten line 5).

Hope it helps!