This question is an exact duplicate of:
- Unexpected Output Running Semaphore 1 answer
The first process shouldn't start its (i)th iteration unless the second process has finished its (i-1)th iteration.The output is not what I need.I wonder if it is possible to have the output by only two semaphores? Here is my code.
#include <stdio.h>
#include <sys/types.h>
#include <unistd.h>
#include <stdlib.h>
#include <errno.h>
#include <semaphore.h>
sem_t semA, semB,sem,m;
int main(void)
{
int i;
pid_t child_a, child_b,pid2,pid3;
sem_init(&semA, 0, 1);
sem_init(&semB, 0, 0);
sem_init(&m, 0, 0);
child_a = fork();
//wait();
if (child_a == 0) {
// int j;
pid2 =getpid();
for (i = 0; i < 5; )
{
sem_wait(&semA);
//sem_wait(&m);
printf("child1: %d\n", i);
i++;
//printf("pid1: %d\n", pid2);
//printf("--------------------------\n");
sleep(3);
//sem_post(&m);
sem_post(&semB);
}
}
else {
child_b = fork();
//wait();
if (child_b == 0) {
pid3 =getpid();
for (i = 0; i < 5;)
{
sem_wait(&semB);
//sem_wait(&m);
printf("child2: %d\n", i);
i++;
//printf("pid2: %d\n", pid3);
//printf("--------------------------\n");
sleep(5);
//sem_post(&m);
sem_post(&semA);
}
}
}
exit(0);
return 0;
}
the output I expect is:
child1: 0
child2: 0
child1: 1
child2: 1
child1: 2
child2: 2
child1: 3
child2: 3
child1: 4
child2: 4
child1: 5
child2: 5
child1: 6
child2: 6
child1: 7
child2: 7
child1: 8
child2: 8
child1: 9
child2: 9
but I get just one child:
child1: 0
(I'm running UBUNTU 12.10)