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 7959705
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: June 4, 20262026-06-04T04:36:55+00:00 2026-06-04T04:36:55+00:00

will a fast enumeration of an array always go in index order? for (NSDictionary

  • 0

will a fast enumeration of an array always go in index order?

for (NSDictionary *zone in myArray)
{
    //do code
}

will that always enumerate from index location 0, 1, 2, …, n?

i prefer fast enumeration over a standard for loop since it makes the code easier to read.

  • 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
    2026-06-04T04:36:56+00:00Added an answer on June 4, 2026 at 4:36 am

    Yes it will:

    For collections or enumerators that have a well-defined order—such as
    an NSArray or an NSEnumerator instance derived from an array—the
    enumeration proceeds in that order, so simply counting iterations
    gives you the proper index into the collection if you need it.

    docs here

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

Sidebar

Related Questions

For getting data from remote url which connection type will work fast URLConnection or
I'm writing a program, that will do some transformations with image from a webcam
I want to have a enumerator/generator that will always provide me with the next
I need a string splitting function that is really fast that will break apart
I'm constructing a database that will grow fast. Some tables will contain several millions
Will the following code: while True: try: print(waiting for 10 seconds...) continue print(never show
Will Java code built and compiled against a 32-bit JDK into 32-bit byte code
Will this code ever wait on the mutex inside the producer's void push(data) ?
Thanks to this question/answer Automatic Reference Counting: Error with fast enumeration I resolved a
I have to store instructions, commands that I will be receiving via serial. 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.