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

  • Home
  • SEARCH
  • 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 3225308
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 17, 20262026-05-17T16:19:28+00:00 2026-05-17T16:19:28+00:00

Using integer math alone, I’d like to safely average two unsigned ints in C++.

  • 0

Using integer math alone, I’d like to “safely” average two unsigned ints in C++.

What I mean by “safely” is avoiding overflows (and anything else that can be thought of).

For instance, averaging 200 and 5000 is easy:

unsigned int a = 200;
unsigned int b = 5000;
unsigned int average = (a + b) / 2; // Equals: 2600 as intended

But in the case of 4294967295 and 5000 then:

unsigned int a = 4294967295;
unsigned int b = 5000;
unsigned int average = (a + b) / 2; // Equals: 2499 instead of 2147486147

The best I’ve come up with is:

unsigned int a = 4294967295;
unsigned int b = 5000;
unsigned int average = (a / 2) + (b / 2); // Equals: 2147486147 as expected

Are there better ways?

  • 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
    2026-05-17T16:19:28+00:00Added an answer on May 17, 2026 at 4:19 pm

    Your last approach seems promising. You can improve on that by manually considering the lowest bits of a and b:

    unsigned int average = (a / 2) + (b / 2) + (a & b & 1);
    

    This gives the correct results in case both a and b are odd.

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

Sidebar

Related Questions

Basically, I have been using both Integer.Parse and CInt in most of my daily
Is it possible to save an integer array using NSUserDefaults on the iPhone? I
How do I programmatically return the maximum of two integers without using any comparison
What is the best way to read an unsigned 24-bit integer from a C#
I often find myself using Integers to represent values in different spaces. For example...
The Dragon Book includes an exercise on converting integers to roman numerals using a
Using online interfaces to a version control system is a nice way to have
Using PyObjC , you can use Python to write Cocoa applications for OS X.
Using ASP.NET MVC there are situations (such as form submission) that may require a
Using C# .NET 3.5 and WCF, I'm trying to write out some of the

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.