Concerning programm semaphore.cpp (program see below) If we change Line #31 as f

Question

Concerning programm semaphore.cpp (program see below)

If we change Line #31 as follows :

... nSemaphoreCount = 3

What will happen with the output? Why?

---------------------

semaphore.cpp

#include <iostream>
#include <pthread.h>       //POSIX thread header
#include <unistd.h>       //UNIX standard header
#include <semaphore.h>
using namespace std;

sem_t g_semt;           //semaphore

void
*fn_work_thread(void *p)
{
   pthread_t tID = pthread_self();       //obtain ID of the calling thread

   cout << "**********Thread " << tID << " is waiting for a semaphore**********" << endl;
   sem_wait(&g_semt);       //decrements the semaphore pointed by g_semt;
   cout << "----------Thread " << tID << " got a semaphore, is running----------" << endl;

   usleep(1000 * 1000 * 2);   //sleep the calling thread, for 2 seconds

   sem_post(&g_semt);       //unlock the semaphore, i.e., increase g_semt with 1

   const char* pRet[] = { "thread finished!" };

   return pRet;
}

int
main(void)
{
   const size_t nThreadCount = 5;       //there are totally 5 threads
   //line 31// const unsigned int nSemaphoreCount = 2;   //initial value of semaphore
   int nRet = -1;
   void *pRet = NULL;
   pthread_t threadIDs[nThreadCount] = { 0 };

   nRet = sem_init(&g_semt, 0, nSemaphoreCount);   //initialize semaphore, here g_semt is set as 2
   if (nRet != 0)
       return -1;

   //start new threads
   for (size_t i = 0; i < nThreadCount; ++i)
   {
       //create new threads
       nRet = pthread_create(&threadIDs[i], NULL, fn_work_thread, NULL);
       if (nRet != 0)
       {
           cout << "Fail to start new thread: " << i << endl;
           continue;
       }
   }

   for (size_t i = 0; i < nThreadCount; ++i)
   {
       int nRet2 = pthread_join(threadIDs[i], &pRet);   //wait for a child thread to terminate
                                                       //   cout << "Thread " << threadIDs[i] << " return value is: " << (char*)pRet << endl;
   }

   cout << endl;
   sem_destroy(&g_semt);       //destroy semaphore

   return 0;
}

Explanation / Answer

If we change nSemaphoreCount as 3 then the output will be

cout<<"----------Thread " << tID << " got a semaphore, is running----------" will be printed 3 times as 3 threads will get semaphore access instead of 2 in the case where nSemaphoreCount was 2.

With each thread access the semaphore will get decremented

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