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 8126823
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: June 6, 20262026-06-06T07:17:40+00:00 2026-06-06T07:17:40+00:00

Possible Duplicate: What does $$ mean in the shell? What are the Special Dollarsign

  • 0

Possible Duplicate:
What does $$ mean in the shell?
What are the Special Dollarsign Variables (Possibly bash Only)?

I am new to linux environment. I am dealing with a particular script that has a command like this

ps -p $$

-p should follow a process id but I don’t understand what $$ means. \
Could some linux gurus shed light on it ? isit a var where is it coming frome etc.,,

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
    2026-06-06T07:17:42+00:00Added an answer on June 6, 2026 at 7:17 am

    The special $$ variable evaluates to the current process’ PID (process ID) — it is an integer number; it’s not listed by env as it’s not an environment variable. It’s a special built-in variable provided by BASH.

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

Sidebar

Related Questions

Possible Duplicate: What does new() mean? Like in title. I wonder what this syntax
Possible Duplicate: What does “(void) new” mean in C++? I'm not familiar with C++
Possible Duplicate: What does it mean “bash < <( curl rvm.io/releases/rvm-install-head )” I'm working
Possible Duplicate: What does $$ mean in the shell? $ ./cruncher & ./cruncher &
Possible Duplicate: What does ||= mean in Ruby? I'm new to ruby and I
Possible Duplicate: C#: What does new() mean? I look at definition of Enum.TryParse: public
Possible Duplicate: What does it mean to have an undefined reference to a static
Possible Duplicate: What does ||= (or equals) mean in Ruby? It's hard to search
Possible Duplicate: What does *args and **kwargs mean? Understanding kwargs in Python I've found
Possible Duplicate: Reference - What does this symbol mean in PHP? I have a

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.