Code to read: #include <stdio.h> #include <stdlib.h> #include <pthread.h> pthrea

Question

Code to read:

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

pthread_mutex_t count_mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_mutex_t count_mutex3 = PTHREAD_MUTEX_INITIALIZER;
int c[2] = {0,0};

void *UpdateC1(void *arg)
{
   int i;
   for(i=0;i<1000000;i++)
   {
       pthread_mutex_lock(&count_mutex);
       c[0]=(c[0]+1)%2;
       c[1]=(c[1]+1)%2;
       pthread_mutex_unlock(&count_mutex);
   }
   return NULL;
}

void *UpdateC2(void *arg)
{
   int i;
   for(i=0;i<1000000;i++)
   {
       pthread_mutex_lock(&count_mutex);
       c[0]=(c[0]+1)%2;
       c[1]=(c[1]+1)%2;
       pthread_mutex_unlock(&count_mutex);
   }
   return NULL;
}

void *UpdateC3(void *arg)
{
   int i;
   for(i=0;i<1000000;i++)
   {
       pthread_mutex_lock(&count_mutex3);
       c[0]=(c[0]+1)%2;
       c[1]=(c[1]+1)%2;
       pthread_mutex_unlock(&count_mutex3);
   }
   return NULL;
}

int main(int argc, char *argv[])
{
   int rt,i;
   pthread_t t[3];

   rt=pthread_create( &t[0], NULL, &UpdateC1, NULL);
   if( rt!=0 )
       fprintf(stderr,"Thread %d creation failed: %d ", 0,rt);
   rt=pthread_create( &t[1], NULL, &UpdateC2, NULL);
   if( rt!=0 )
       fprintf(stderr,"Thread %d creation failed: %d ", 1,rt);
   rt=pthread_create( &t[2], NULL, &UpdateC3, NULL);
   if( rt!=0 )
       fprintf(stderr,"Thread %d creation failed: %d ", 2,rt);

   for(i=0;i<3;i++)
   {
       rt=pthread_join( t[i], NULL);
       if( rt!=0 )
           fprintf(stderr,"Wait for thread %d failed: %d ", i,rt);
   }

   printf (" %c %c ",'a'+c[0],'a'+c[1]);
   return 0;
}

1. Read given code RaceOrNot1.c and write all possible outputs of the program. Assume there will be no thread creation or joining failures or mutex failures. If you believe there is only one possible output, you just need to write that output

Explanation / Answer

Possible outputs are:

a a

a b

b a

exaplaination:

at any time one thread will be running as there is a mutex lock wheneven there is a updation in each thread execution. So 3 possible values for the array of C are (0,0) , (0,1) , (1,0)

After the all theards completes their execution , there is printf which prints the a character

a + c[0] a+c[1] ie a + 0 a+ 0=> a a

a + c[0] a+c[1] ie a +0 , a+1 => a b

a + c[0] a+c[1] ie a +1 a+0 => b a

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