About assignment by reference and value in JAVA

Question

Case 1

public static void main(String[] args) {
    Dog aDog = new Dog()
    Dog oldDog = new Dog()
if(aDog== oldDog) {
  System.out.println("Same");
} else {
  System.out.println("Different");
}

}

Output: Different

Case 2

public static void main(String[] args) {
    Dog aDog = new Dog();
    Dog oldDog = aDog;

if(aDog== oldDog) {
  System.out.println("Same");
} else {
  System.out.println("Different");
}

}

Output: Same

Can someone explain to me the result of these programs, please?

Answer 1

• The first case creates two different Objects, each one with its own reference/entity in memory. Instances from the same Class, but different Objects anyway.
• In the second case, you assign the same Object reference to the new Dog instance: they both refer to the same Object.

The == operator compares the references of objects. In your second case, they both reference the same Dog.

This is for Objects, not primitives; Regarding primitives, the value itself is compared.

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Java Specification Equality Operators:

If the operands of an equality operator are both of either reference type or the null type, then the operation is object equality. A compile-time error occurs if it is impossible to convert the type of either operand to the type of the other by a casting conversion (§5.5). The run-time values of the two operands would necessarily be unequal.

At run time, the result of == is true if the operand values are both null or both refer to the same object or array; otherwise, the result is false.