MultiThreading issues with synch

Question

1.00/5 (1 vote)

See more:

Hi There,

I have created for 10 threads using AfxbeginThread API.

but i wants to start all 10 threads at a time, i don't wants to start the thread one by one.

Can any one can help how to implement.

Regards,
Ranjith

Posted 22-Jan-13 1:31am

ranjithkumar81

Add a Solution

Comments

Sergey Alexandrovich Kryukov 22-Jan-13 11:18am

This is not a question, and you statement simply makes no sense. If you think about it, you will see that there is no difference.
—SA

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

chaau · Accepted Answer · 2013-01-22T12:09:00

It is very easy to do. In the main program body (main thread) create an Event Object[^] with the initial state nonsignaled. Pass the event handle to all your threads. Then in all the thread routines specify WaitForSingleObject() to wait for the event to become signaled. Once you have created all the threads, put the event to signaled state using SetEvent(). All your threads will continue execution.