how to context switch to have a result at the end of each running

Question

0.00/5 (No votes)

See more:

C++

/*

 Add 10 to a variable in each of two processes.

 The answer can be between 2 and 20.

 Local variable enables bad scenario with source-level interleaving.

*/

int sum = 0;

void add10() {
	int i;
	int local;
	for (i = 1; i <= 10;i++)
 {
		local = sum;
		sum = local + 1;
}
}

void main() {
	cobegin { 
		add10(); 
                add10();
        }
        cout << "Sum = " << sum << endl;
}

.....................................................
I use jbaci(is a program that run functioد that are in "cobegin" coincidence

but each time that I run this I see different results between 0. and 20

I dont know how to solve this prob with semaphore or other method

please help me

thank you for decision for solve this

Posted 30-Oct-10 7:35am

afshin.b.k

Updated 30-Oct-10 21:32pm

v6

Add a Solution

Comments

Dalek Dave 30-Oct-10 17:59pm

Edited for Grammar and Readability.

2 solutions

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)

Nish Nishant · Answer 1 · 2010-10-30T07:52:00

Solution 1

I have not used jbaci, but assuming it's some kind of parallel environment/library, you need to make sure you only let one thread write to sum (via some kind of synchronized block, jbaci should have one I assume).

See http://www.codeproject.com/KB/threads/lwsync.aspx[^]

Posted 30-Oct-10 7:52am

Nish Nishant

Updated 30-Oct-10 7:54am

v2

Comments

Dalek Dave 30-Oct-10 17:59pm

Good old CP Archives!

afshin.b.k · Answer 2 · 2010-10-31T03:43:00

/* Add 10 to a variable in each of two processes.
The answer can be between 2 and 20.
Local variable enables bad scenario with source-level interleaving.
*/

int sum = 0;

int i;
int local;
semaphore w=1;

void add10(int f)
{

wait(w);

for(i=1;i<=10;i++){local=sum;sum=local+1;}

signal(w);

}

void main() {
cobegin {
add10(0);
add10(1);
}
cout << "Sum = " << sum << endl;
}