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 3695598
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 19, 20262026-05-19T04:38:14+00:00 2026-05-19T04:38:14+00:00

If I want to instantiate a LinkedList and require access to the methods in

  • 0

If I want to instantiate a LinkedList and require access to the methods in both List and Dequeue interfaces, and do not want to type to the concrete implementation, and do not want to cast between interfaces, is there a way?

i.e.:

LinkedList ll = new LinkedList(); // don't want to do this...


List ll = new LinkedList();
ll.peekFirst(); // can't access peekFirst method
((DeQueue) ll).peekFirst(); // Kinda ugly
  • 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
    2026-05-19T04:38:14+00:00Added an answer on May 19, 2026 at 4:38 am
    public interface Foo<T> extends List<T>, Deque<T>{}
    
    public class Bar<T> extends LinkedList<T> implements Foo<T>{}
    
    Foo ll = new Bar();
    

    Don’t have eclipse near me, this should compile though.

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

Sidebar

Related Questions

Because not all the classes used in the class I want to instantiate are
let's say I want to instantiate a different type of object depending on certain
I want to instantiate 2 property pages from the same class and template, because
I have a python class I want to instantiate and the __init__ definition has
I have custom classes that I currently instantiate within App.xaml as resources. I want
Is this possible? What I want to do is have my doWork method instantiate
I have a class that instantiates two classes which implement interfaces. I want one
want to know why String behaves like value type while using ==. String s1
I want to initialize a linked list with pointer arguments like so: /* *
I want to instantiate a Vector.<Number> with two values of 1.0 You might think

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.