Thanks all for the response.
I thought of constructing a conditional wait using semaphores as follows:
//global flag = 0.
void *text(void *arg)
{
int n = *(int*)arg;
while(flag != n)
{
sem_wait(mutex);
internal_flag = 1;
}
if(internal_flag)
sem_post(mutex);
Folks, I have one pthread question.I know that its not the right place but
i thought this is the right place to post this. Code snippet:
#include stdlib.h
#include stdio.h
#include unistd.h /* defines _POSIX_THREADS if pthreads are available */
#ifdef _POSIX_THREADS
# include pthread.h
Hi,
As you are creating a thread and soon after you are joining for it.
{
pthread_create(tid[i], NULL, text, (void*)code[i]);
pthread_join(tid[i],NULL);
}
then all the threads wont run in parallel.
Like thread[1] will get started and then main thread will wait for it.
Do you really want to do
You can use mutex instead of semaphores if you need to execute only one
case at a time.
FROM,
V.VIGNESH.
M.Sc Theoretical Computer Science.
PSG College of Technology
On 25 November 2012 22:37, jagannath jpdasi...@gmail.com wrote:
Folks, I have one pthread question.I know that its not the
@Naveen: I didn't understand your point . Can you please throw more light
on join operation
On Sun, Nov 25, 2012 at 10:47 PM, vIGNESH v v.v.07121...@gmail.com wrote:
You can use mutex instead of semaphores if you need to execute only one
case at a time.
FROM,
V.VIGNESH.
M.Sc Theoretical
According to your code snippet, all threads would execute one by one
serially i.e. first thread would be created, would execute text routine
and terminate; then second thread would be created, would execute text
routine and terminate and so on. Hence, multiple threads are not running in
parallel.
