Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 6194351
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 24, 20262026-05-24T03:16:55+00:00 2026-05-24T03:16:55+00:00

I have a custom class called Position. I want to use the following:- Set<Position>

  • 0

I have a custom class called Position. I want to use the following:-

Set<Position> s=new HashSet<Position>();

Do I have to override the HashCode() method in the Position class? I have not overridden equals() method. I want two Position objects to be considered equal only when they are the same object. Do I still have to override HashCode() in order to use the HashSet as shown above?

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-05-24T03:16:55+00:00Added an answer on May 24, 2026 at 3:16 am

    You only have to override hashcode and equals if you want different objects (presumably representing the same thing) to count as equal.

    (To clarify, if you do override one you should override the other, so that objects that test as equal always have the same hashcode.)

    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

I have a custom class called Rows that implements IEnumerable<Row> . I often use
I have a custom class called NoteView that extends FrameLayout . Previously, I had
I have a situation where I want to use a custom EditorTemplate with a
I have a custom class loader so that a desktop application can dynamically start
I have a custom class that implements that IComparable. This class is stored in
I have a custom class that implements ICollection , and this class is readonly,
I have a custom class Contact . I am trying to bind a List<Contact>
Say you have a custom class call it Foo. When you then have an
I currently have a custom session handler class which simply builds on php's session
I have a std::multimap where key is a custom class. Something like this: Class

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.