Posix threads synchronisation with semaphores

[evariste]

Hi all,

can someone explain for me the output of this program :

#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 :

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

and run it :

./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.

[cas]

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

[evariste]

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

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.

[Codeplug]

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

[evariste]

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.

[Codeplug]

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

[evariste]

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

Regards.