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 605321
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 13, 20262026-05-13T17:07:57+00:00 2026-05-13T17:07:57+00:00

I will listen on a port (simple server) when a request is passed parse

  • 0

I will listen on a port (simple server) when a request is passed parse the URL
and start a thread.

The thread will insert an element in a queue which is shared, and it is locked while inserting.

I am not able to get element when I call peek on queue.

use Thread qw(async); 
use Thread::Queue;

 my $DataQueue:shared = new Thread::Queue; 

 $newElement = init($user,$param,$reqest);  # init is method in ElementStructure.pm
 #after creating the element it is passes to subroutine where thread is started

sub updateData
{
    my $iElement=shift;

    $thr = async 
    { 

        {
            lock($DQueue);

            print "---->locked\n";
                    $DQueue->enqueue($iElement);
            insertdata();

        }

        print "lock released\n";

    };
}

sub insertdata
{
     my $count=0;
     while ($DataElement = $DQueue->peek($count) )
     {
    print "-- position $count\n";
    $count++;
     }
}
  • 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
    2026-05-13T17:07:58+00:00Added an answer on May 13, 2026 at 5:07 pm

    Perhaps the problem is that you use $DataQueue one place but $DQueue elsewhere? Make sure you are using strict and warnings.

    If $iElement may be false (e.g. 0), you will need to say

    while ( defined ( my $DataElement = $DQueue->peek($count) ) )
    

    Correcting the variable name and putting in some code to call updateData, everything seemed to work for me.

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

Sidebar

Ask A Question

Stats

  • Questions 373k
  • Answers 373k
  • Best Answers 0
  • User 1
  • Popular
  • Answers
  • Editorial Team

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

    • 7 Answers
  • Editorial Team

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

    • 5 Answers
  • Editorial Team

    What is a programmer’s life like?

    • 5 Answers
  • Editorial Team
    Editorial Team added an answer The following, found on the internet, addressed my dilemma. As… May 14, 2026 at 7:38 pm
  • Editorial Team
    Editorial Team added an answer mod_rewrite will match things in braces (), and then you… May 14, 2026 at 7:38 pm
  • Editorial Team
    Editorial Team added an answer I believe what you are looking for is an Anti-Cross… May 14, 2026 at 7:38 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.