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 3308180
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 17, 20262026-05-17T21:29:31+00:00 2026-05-17T21:29:31+00:00

Quoting from Code Complete 2 , int Factorial( int number ) { if (

  • 0

Quoting from Code Complete 2,

int Factorial( int number ) {
   if ( number == 1 ) {
      return 1;
   }
   else {
      return number * Factorial( number - 1 );
   }
}

In addition to being slow [1] and making
the use of run-time memory
unpredictable
[2], the recursive version
of this routine is harder to
understand than the iterative version,
which follows:

int Factorial( int number ) {
   int intermediateResult = 1;
   for ( int factor = 2; factor <= number; factor++ ) {
      intermediateResult = intermediateResult * factor;
   }
   return intermediateResult;
}

I think the slow part is because of the unnecessary function call overheads.

But how does recursion make the use of run-time memory unpredictable?

Can’t we always predict how much memory would be needed (as we know when the recursion is supposed to end)? I think it would be as unpredictable as the iterative case, but not any more.

  • 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
    2026-05-17T21:29:32+00:00Added an answer on May 17, 2026 at 9:29 pm

    Because of the fact recursive methods call them selves repeatedly, the need lots of stack memory. Since the stack is limited, errors will occur if the stack memoy is exceeded.

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

Sidebar

Related Questions

Quoting from the MSDN Link for ValueType Class In cases where it is necessary
I'm quoting part of an answer which I received for another question of mine
I'm quoting myself on a previous question I asked to further explain: I'm trying
Which style of Ruby string quoting do you favour? Up until now I've always
For some reasons, I would like to do an explicit quoting of a string

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.