在C Linux中使用三个线程使用信号量同步顺序打印3 4 5 50次

Question

My programming is going in dead lock.I am trying to print three numbers 3 4 5 sequentially for 50 times using three threads using semaphore synchronization. Please help me.

Below is the code

#include <iostream>
#include <pthread.h>
#include <semaphore.h>

using namespace std;

sem_t sem1;
sem_t sem2;
sem_t sem3;

void * fun1(void *)
{   
    for(int i = 0; i < 50 ; i++)
    {
    sem_wait(&sem1);
    sem_wait(&sem3);
    cout<<"3"
    sem_post(&sem2);
    sem_post(&sem3);
    }
}


void * fun2(void *)
{   

   for(int i = 0; i < 50 ; i++)
   {
    sem_wait(&sem2);
    sem_wait(&sem3);
    cout<<"4";
    sem_post(&sem3);
    sem_post(&sem1);
   }

}

void * fun3 (void *)
{
   for(int i = 0; i< 50; i++)
   {

    sem_wait(&sem2);
    sem_wait(&sem3);
    cout<<"5";
    sem_post(&sem1);
    sem_post(&sem2);

   }
}

int main()
{
   pthread_t t1;
   pthread_t t2;
   pthread_t t3;


   sem_init(&sem1,0,1);
   sem_init(&sem2,0,0);
   sem_init(&sem3,0,1);

   pthread_create(&t1,NULL,&fun1,NULL);
   pthread_create(&t2,NULL,&fun2,NULL);
   pthread_create(&t3,NULL,&fun3,NULL); 

   pthread_join(t1,NULL);
   pthread_join(t2,NULL);
   pthread_join(t3,NULL);

   return 1;
}

Please help me to understand and solve this deadlock.Provide suggestions also i can do this for example 3 4 5 6 using 4 etc threads

Answer 1

Please help me to understand and solve this deadlock.

There is indeed a deadlock in your code. Consider at the beginning, thread 1 first gets 2 semaphores and call cout << "3" . After posting sem2 and sem3 , it is possible that thread 3 immediately gets these 2 sem, then call cout << "5" . However, after thread 3 posting sem1 and sem2 , no one can reach a cout << statement, because sem3 's value is 0 and everyone needs to pass a wait of sem3 .

If you are wondering why there is totally no output, it's because the buffer inside iostream . For console output, "\\n" will flush buffer, so if you replace "3" by "3\\n" , you can see the output.

Provide suggestions also i can do this for example 3 4 5 6 using 4 etc threads

In the following code, you should see the symmetry, which can be easily generalized to any number of thread. And you should always call sem_destroy after using semaphore, otherwise you might get system level resource leak.

#include <iostream>
#include <pthread.h>
#include <semaphore.h>

using namespace std;

sem_t sem1;
sem_t sem2;
sem_t sem3;

void * fun1(void *)
{   
    for(int i = 0; i < 50 ; i++)
    {
    sem_wait(&sem1);
    cout<<"3\n";
    sem_post(&sem2);
    }
}


void * fun2(void *)
{   

   for(int i = 0; i < 50 ; i++)
   {
    sem_wait(&sem2);
    cout<<"4\n";
    sem_post(&sem3);
   }

}

void * fun3 (void *)
{
   for(int i = 0; i< 50; i++)
   {

    sem_wait(&sem3);
    cout<<"5\n";
    sem_post(&sem1);

   }
}

int main()
{
   pthread_t t1;
   pthread_t t2;
   pthread_t t3;


   sem_init(&sem1,0,1);
   sem_init(&sem2,0,0);
   sem_init(&sem3,0,0);

   pthread_create(&t1,NULL,&fun1,NULL);
   pthread_create(&t2,NULL,&fun2,NULL);
   pthread_create(&t3,NULL,&fun3,NULL); 

   pthread_join(t1,NULL);
   pthread_join(t2,NULL);
   pthread_join(t3,NULL);

   sem_destroy(&sem1);
   sem_destroy(&sem2);
   sem_destroy(&sem3);

   return 1;
}