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 3666544
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 19, 20262026-05-19T01:53:36+00:00 2026-05-19T01:53:36+00:00

Say I have: NSDictionary *stuff; // {1 => hi, 2 => bye} NSArray *array

  • 0

Say I have:

NSDictionary *stuff; // {"1" => "hi", "2" => "bye"}
NSArray *array = [stuff allKeys];

allKeys makes a copy of stuff’s keys, so array is now responsible for releasing this information.

Later on, when I want to

I cannot do:

array = [newStuff allKeys];

because it would just reassign the pointers and orphan the original array. I must first remove the objects

[array removeAllObjects];

Wanted to know if my understand is correct? Thanks!

  • 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
    2026-05-19T01:53:36+00:00Added an answer on May 19, 2026 at 1:53 am

    Not quite.

    NSArray *array = [stuff allKeys];
    

    This gives you an array that you don’t own. Whether it’s technically a copy or not is not your problem. Since the accessor doesn’t start with the word “alloc” or “new”, or contain the word “copy”, you don’t own the return value, which means you don’t need to release it. (But you do need to retain it if you want to keep it.)

    If you later do this:

    array = [newStuff allKeys];
    

    that’s fine. It stomps on the original reference, as you know, but since you don’t own that reference anyways, it’s OK to let it go. This new reference is also, of course, not yours unless you retain it.

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

Sidebar

Related Questions

Say I have three files (template_*.txt): template_x.txt template_y.txt template_z.txt I want to copy them
I have an NSDictionary where each key points to an array. I later want
Let's say I have an NSArray of NSDictionaries that is 10 elements long. I
Let's say have something like: SELECT energy_produced, energy_consumed, timestamp1 AS timestamp FROM ( SELECT
Say we have the following method: private MyObject foo = new MyObject(); // and
Say I have the following file structure: app/ app.py controllers/ __init__.py project.py plugin.py If
Say you have an application divided into 3-tiers: GUI, business logic, and data access.
Say I have an ASMX web service, MyService. The service has a method, MyMethod.
Say I have a class named Frog, it looks like: public class Frog {
Say I have two strings, String s1 = AbBaCca; String s2 = bac; I

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.