DZone
Thanks for visiting DZone today,
Edit Profile
  • Manage Email Subscriptions
  • How to Post to DZone
  • Article Submission Guidelines
Sign Out View Profile
  • Post an Article
  • Manage My Drafts
Over 2 million developers have joined DZone.
Log In / Join
Refcards Trend Reports
Events Video Library
Refcards
Trend Reports

Events

View Events Video Library

Related

  • Top Java Security Vulnerabilities and How to Prevent Them in Modern Java
  • A Spring Boot App With Half the Startup Time
  • Implementing the Planning Pattern With Java Enterprise and LangChain4j
  • Getting Started With Agentic Workflows in Java and Quarkus

Trending

  • The Agentic Agile Office: Streamlining Enterprise Agile With Autonomous AI Agents
  • Testing AI-Infused Apps: A Dual-Layer Framework for AI Quality Assurance
  • Grok AI API Tutorial: Chat, Image, Video, Tool Calling, and Web Search
  • Persistent Memory for AI Agents Using LangChain's Deep Agents
  1. DZone
  2. Coding
  3. Java
  4. How Does Java Handle Aliasing?

How Does Java Handle Aliasing?

By 
Ryan Wang user avatar
Ryan Wang
·
Mar. 30, 13 · Interview
Likes (0)
Comment
Save
Tweet
Share
37.2K Views

Join the DZone community and get the full member experience.

Join For Free

Aliasing means there are multiple aliases to a location that can be updated, and these aliases have different types.

In the following example, a and b are two variable names that have two different types A and B. B extends A.

B[] b = new B[10];
A[] a = b;
 
a[0] =  new A();
b[0].methodParent();

In memory, they both refer to the same location.

Java Aliasing

The pointed memory location are pointed by both a and b. During run-time, the actual object stored determines which method to call.

How does Java handle aliasing problem?

If you copy this code to your eclipse, there will be no compilation errors.

class A {
	public void methodParent() {
		System.out.println("method in Parent");
	}
}

class B extends A {
	public void methodParent() {
		System.out.println("override method in Child");
	}

	public void methodChild() {
		System.out.println("method in Child");
	}
}

public class Main {
	public static void main(String[] args) {
		B[] b = new B[10];
		A[] a = b;
 
		a[0] =  new A();
		b[0].methodParent();
	}
}

But if you run the code, the output would be as follows:

Exception in thread “main” java.lang.ArrayStoreException: aliasingtest.A
at aliasingtest.Main.main(Main.java:26)

The reason is that Java handles aliasing during run-time. During run-time, it knows that the first element should be a B object, instead of A.

Therefore, it only runs correctly if it is changed to:

B[] b = new B[10];
A[] a = b;

a[0] =  new B();
b[0].methodParent();

and the output is:

override method in Child
* original article 
Java (programming language)

Published at DZone with permission of Ryan Wang. See the original article here.

Opinions expressed by DZone contributors are their own.

Related

  • Top Java Security Vulnerabilities and How to Prevent Them in Modern Java
  • A Spring Boot App With Half the Startup Time
  • Implementing the Planning Pattern With Java Enterprise and LangChain4j
  • Getting Started With Agentic Workflows in Java and Quarkus

Partner Resources

×

Comments

The likes didn't load as expected. Please refresh the page and try again.

  • RSS
  • X
  • Facebook

ABOUT US

  • About DZone
  • Support and feedback
  • Community research

ADVERTISE

  • Advertise with DZone

CONTRIBUTE ON DZONE

  • Article Submission Guidelines
  • Become a Contributor
  • Core Program
  • Visit the Writers' Zone

LEGAL

  • Terms of Service
  • Privacy Policy

CONTACT US

  • 3343 Perimeter Hill Drive
  • Suite 215
  • Nashville, TN 37211
  • [email protected]

Let's be friends:

  • RSS
  • X
  • Facebook