In Java, how do you assign the address (reference) of one object to another reference variable of the same type?

Introduction / Context:
In Java, variables that refer to objects store references or addresses, not the objects themselves. Sometimes you want another variable to refer to the same object so that changes through either reference are visible. Understanding how to assign references is fundamental to working with objects and collections. Interviewers may ask about this to see whether you understand the difference between copying a reference and copying object state.

Given Data / Assumptions:

• We have a reference variable ref1 of some class type RefType that already refers to a valid object.
• We want a second variable ref2 to refer to the same underlying object as ref1.
• We are not performing a deep copy of the object's fields, only copying the reference value.
• Both ref1 and ref2 are of the same reference type or compatible types in the class hierarchy.

Concept / Approach:
In Java, you assign references using the same assignment operator that you use for primitives. When you write RefType ref2 = ref1;, the value stored in ref1, which is the reference to the object on the heap, is copied into ref2. After the assignment, both variables refer to the same object. Any changes made to the object through one reference are visible when accessed through the other reference. No new object is created by this assignment; only the reference value is copied. This is similar to copying a pointer value in pointer based languages, but Java hides pointer syntax and uses reference semantics instead.

Step-by-Step Solution:
Step 1: Ensure that ref1 already refers to an object, for example RefType ref1 = new RefType();. Step 2: Declare another reference variable of a compatible type, such as RefType ref2;. Step 3: Use assignment with the equals operator to copy the reference value: ref2 = ref1;. Step 4: After this statement, both ref1 and ref2 point to the same object on the heap. Step 5: Verify that modifying the object through ref2, such as ref2.setValue(10);, is visible when you access it through ref1, confirming that the underlying object is shared.

Verification / Alternative check:
Writing a small test program shows this behavior clearly. Create a simple class with a field and a setter method, assign ref2 = ref1;, change the field through ref2, and then print it through ref1. The output will show the updated value, proving that both references refer to the same object. Using a debugger also reveals that the memory address or identity hash codes associated with ref1 and ref2 are identical after assignment, confirming that the reference was copied, not the object state.

Why Other Options Are Wrong:
Option B is incorrect because there is no built in assignAddress statement or method in Java for copying references; simple assignment covers this use case. Option C is wrong because Java does not have a copy keyword; copying references uses the standard assignment operator. Option D is clearly incorrect since assigning one reference to another is a basic operation and is widely used throughout Java code, especially in methods that return or cache references.

Common Pitfalls:
A common pitfall is assuming that assigning one reference to another creates a deep copy of the object. In reality, only the reference value is copied, so both variables refer to the same object. This can lead to unintended side effects when mutating shared objects. Another mistake is forgetting to handle null; assigning a null reference to another variable is legal but can cause NullPointerException later if methods are called without proper checks. Developers should be clear about whether they want shared references or independent copies of data structures and use cloning or copy constructors when they truly need deep copies.

Final Answer:
To assign the address of one object to another reference variable in Java, you use simple assignment, for example RefType ref2 = ref1;, which copies the reference value so that both variables point to the same object.