Posix threads synchronisation with semaphores

Hi all,

can someone explain for me the output of this program :

Code:

---

#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <string.h>
#include <pthread.h>
#include <semaphore.h>

void *thread_function(void *arg);
sem_t bin_sem;

#define WORK_SIZE 1024
char work_area[WORK_SIZE];

int main() {
  int res;
  pthread_t a_thread;
  void *thread_result;

  res = sem_init(&bin_sem, 0, 0);
  if (res != 0) {
      perror("Semaphore initialization failed");
      exit(EXIT_FAILURE);
  }
  res = pthread_create(&a_thread, NULL, thread_function, NULL);
  if (res != 0) {
      perror("Thread creation failed");
      exit(EXIT_FAILURE);
  }

  printf("Input some text. Enter 'end' to finish\n");
  while(strncmp("end", work_area, 3) != 0) {
    if (strncmp(work_area, "FAST", 4) == 0) {
      sem_post(&bin_sem);
      strcpy(work_area, "Wheeee...");
    } else {
      fgets(work_area, WORK_SIZE, stdin);
    }
    sem_post(&bin_sem);
  }

  printf("\nWaiting for thread to finish...\n");
  res = pthread_join(a_thread, &thread_result);
  if (res != 0) {
      perror("Thread join failed");
      exit(EXIT_FAILURE);
  }
  printf("Thread joined\n");
  sem_destroy(&bin_sem);
  exit(EXIT_SUCCESS);
}

void *thread_function(void *arg) {
  sem_wait(&bin_sem);
  while(strncmp("end", work_area, 3) != 0) {
      printf("You input %d characters\n", strlen(work_area) -1);
      sem_wait(&bin_sem);
  }
  pthread_exit(NULL);
}

---

when you compile :

Code:

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gcc -Wall -D_REENTRANT thread1.c -o thread1 -lpthread

---

and run it :

Code:

---

./thread1
Input some text. Enter 'end' to finish
hello
You input 5 characters
i m a thread
You input 12 characters
FAST
You input 8 characters
You input 8 characters
You input 8 characters
end

Waiting for thread to finish...
Thread joined

---

and this is not clear for me.

Thanks for your help.

Regards.

In the future, please explain what is not clear. It's good that you showed the output of your program, but you did not explain why it is incorrect. I can only assume that you're confused about the fact that FAST results in "You printed.." being shown 3 times, but I really don't know.

I'll have to assume that's the case. When you type FAST, you cause sem_post() to be called three times, which causes sem_wait() to wake up three times.

a) You type in FAST: the sem_post() after your if/else block is called.
b) Your strncmp() (which could just be a strcmp()) matches FAST, so sem_post() in that block is called.
c) The sem_post() from (a) is called again

Thanks cas for the reply, i'm sorry about explaining what is not clear. But i'm very confused about this output:

Code:

---

FAST
You input 8 characters
You input 8 characters
You input 8 characters

---

I was expecting just one time this sequence but i got it three times so it's not clear at all.

Thanks again.

Regards.

Officially, you have undefined behavior due to unsynchronized access to "work_area".

>> I was expecting just one ... but i got it three times ...
You see the message 3 times because main() called sem_post() 3 times. cas identified all 3 calls in "a) b) c)" of post #2.

gg

Thanks for your reply Codeplug but i don't understand the call in a) and c) from the reply of cas? When i type FAST i increment the semaphore so the thread_function executes and ouput 3!!! right?

Thanks.

1) "FAST" is entered, fgets() returns, POST
2) loop to top of while loop, work_area == "FAST", POST
3) bottom of while loop is reached, POST

There are your 3 sem_post() calls.

4) loop to top of while loop, call fgets() and wait for input

In the meantime, thread_function() processes all 3 sem_post() calls, and see's "Wheeee...".

gg

That's much better :), thank you very much codeplug.

Regards.