Question

Read given code RaceOrNot3.c and write all possible outputs of the program. Assume there will be no thread creation or joining failures or semaphore failures. If you believe there is only one possible output, you just need to write that output.

#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <semaphore.h>
#include <errno.h>

sem_t s1;
int c=0,x=0;

void *UpdateC1(void *arg)
{
   int i;
   for(i=0;i<2000000;i++)   {
       sem_wait(&s1);
       c++;   x++;
       sem_post(&s1);
   }
}

void *UpdateC2(void *arg)
{
   int i,x=0;
   for(i=0;i<2999999;i++)   {
       sem_wait(&s1);
       c++;   x++;
       sem_post(&s1);
   }
}

int main(int argc, char *argv[])
{
   int rt,i;
   pthread_t t[2];

   if(sem_init(&s1, 0, 1)==-1)//Initialize the semaphore;
   {
       fprintf(stderr,"sem_init failed. errno=%d\n",errno);
       exit(1);
   }
   rt=pthread_create( &t[0], NULL, UpdateC1, NULL);
   if( rt!=0 )
       fprintf(stderr,"Thread %d creation failed: %d\n", 0,rt);
   rt=pthread_create( &t[1], NULL, UpdateC2, NULL);
   if( rt!=0 )
       fprintf(stderr,"Thread %d creation failed: %d\n", 1,rt);

   for(i=0;i<2;i++)
   {
       rt=pthread_join(t[i], NULL);
       if( rt!=0 )
           fprintf(stderr,"Wait for thread %d failed: %d\n", i,rt);
   }

   printf ("\t%c\t%c\n",'b'-(c%2),'b'-(x%2));
   return 0;
}

Answer 1

There is only one possible out for this program it is:

a       b

This is because, the variable x which is global, is incremented in thread C1 only, and at the end of this thread the value of x is 2000000, as this variable is incremented by 1 for 2000000 times in the for loop. The variable x used in thread C2 is a local variable which is visible in that thread alone. Thus the variable x used in the main function in printf statement take the value of x which is updated in thread C1.

Hence x%2 is 0 as 2000000%2 is 0. Therefore 'b' - (x%2) will be 'b' - 0 which will be same as character 'b'.

Whereas the variable c is a global variable and both the threads C1 and C2 modify this variable. Thus the variable c is incremented by 1 for 2000000 times in for loop in thread C1 and it is incremented by 1 for 2999999 times in for loop in thread C2, thus in total 2000000 + 2999999 = 4999999 times incremented by 1. Therefore when both the threads finished their execution, the value of variable c is 4999999. As this variable is global, it is visible in main function with the value 4999999.

Hence c%2 is 1 as 4999999%2 is 1. Therefore 'b' - (c%2) will be 'b' - 1 which will be the character 'a'.

Thus the output:    a       b