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 761767
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 14, 20262026-05-14T15:52:40+00:00 2026-05-14T15:52:40+00:00

It seems that in language like C, Java, and Ruby (as opposed to Javascript),

  • 0

It seems that in language like C, Java, and Ruby (as opposed to Javascript), a new scope is created for each iteration of a loop block, and the local variable defined for the loop is actually made into a local variable every single time and recorded in this new scope?

For example, in Ruby:

p RUBY_VERSION

$foo = []

(1..5).each do |i|
  $foo[i] = lambda { p i }
end

(1..5).each do |j|
  $foo[j].call()
end

the print out is:

[MacBook01:~] $ ruby scope.rb
"1.8.6"
1
2
3
4
5
[MacBook01:~] $ 

So, it looks like when a new scope is created, a new local copy of i is also created and recorded in this new scope, so that when the function is executed at a later time, the “i” is found in those scope chains as 1, 2, 3, 4, 5 respectively. Is this true? (It sounds like a heavy operation).

Contrast that with

p RUBY_VERSION

$foo = []

i = 0

(1..5).each do |i|
  $foo[i] = lambda { p i }
end

(1..5).each do |j|
  $foo[j].call()
end

This time, the i is defined before entering the loop, so Ruby 1.8.6 will not put this i in the new scope created for the loop block, and therefore when the i is looked up in the scope chain, it always refer to the i that was in the outside scope, and give 5 every time:

[MacBook01:~] $ ruby scope2.rb
"1.8.6"
5
5
5
5
5
[MacBook01:~] $ 

I heard that in Ruby 1.9, i will be treated as a local defined for the loop even when there is an i defined earlier?

The operation of creating a new scope, creating a new local copy of i each time through the loop seems heavy, as it seems it wouldn’t have matter if we are not invoking the functions at a later time. So when the functions don’t need to be invoked at a later time, could the interpreter and the compiler to C / Java try to optimize it so that there is no local copy of i each time?

  • 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
    2026-05-14T15:52:40+00:00Added an answer on May 14, 2026 at 3:52 pm

    This is similar to the topic discussed here: http://math.andrej.com/2009/04/09/pythons-lambda-is-broken/ . In some programming languages, when you loop over a variable, the body of the loop construct is bound to one variable that increments. In others, a new loop variable is instantiated per iteration of the loop.

    As for lexical scope, note that in JavaScript, functions are the only constructs that form scopes (braces for if, while, for, etc. do not). In C/C++, any pair of braces forms a scope.

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

Sidebar

Ask A Question

Stats

  • Questions 434k
  • Answers 434k
  • Best Answers 0
  • User 1
  • Popular
  • Answers
  • Editorial Team

    How to approach applying for a job at a company ...

    • 7 Answers
  • Editorial Team

    What is a programmer’s life like?

    • 5 Answers
  • Editorial Team

    How to handle personal stress caused by utterly incompetent and ...

    • 5 Answers
  • Editorial Team
    Editorial Team added an answer You don't need jquery for that, you can use an… May 15, 2026 at 3:25 pm
  • Editorial Team
    Editorial Team added an answer If all the methods in your controller need the same… May 15, 2026 at 3:25 pm
  • Editorial Team
    Editorial Team added an answer Given the way you've asked multiple questions you probably realize… May 15, 2026 at 3:25 pm

Trending Tags

analytics british company computer developers django employee employer english facebook french google interview javascript language life php programmer programs salary

Top Members

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.