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 8854559
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: June 14, 20262026-06-14T13:54:39+00:00 2026-06-14T13:54:39+00:00

Just a quick question about the bitwise operator &. If know that x &

  • 0

Just a quick question about the bitwise operator &.

If know that x & y == z and I know the value of y and z is there a way I can calculate the value of x?

  • 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
    2026-06-14T13:54:41+00:00Added an answer on June 14, 2026 at 1:54 pm

    If you are told that x is unique, then no, there is no way of doing it for arbitrary values of y. Otherwise, there are 2n different solutions to that equation, where n is the number of zero bits of y.

    As an example, let’s consider single bit numbers. If y is one, then the value of x must be the same as z (1&1 = 1; 0&1 = 0). If y is zero, z will also be zero, and x can be either one or zero (0&0 = 0; 1&0 = 0).

    On many-bit numbers, every zero bit on y doubles the number of possible values of x, hence 2n

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

Sidebar

Related Questions

Just a quick question about best practice in MVC development. Let's say that I've
Just a quick question about hover events, How can I send a hover event
I just have a quick question about the conditional operator. Still a budding programmer
Just a quick question about a function like this: class Test { public: Test(vector<int>&
Just a quick question about ReadWriteLocks in Java (specifically the ReentrantReadWriteLock implementation) as I
Just a quick question about how parameters are passed in Java... ... if ((index
Just a quick question about how you would go about implementing this. I want
Just a quick question about using select() . I'm using select() to read from
Hey guys just a quick question about the performance of drawRect: as I've noticed
I just had a quick question about git - how backwards compatible are git

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.