Write a program that is similar to cond1.c except that: thread1: increments coun
ID: 3862257 • Letter: W
Question
Write a program that is similar to cond1.c except that:
thread1: increments count only when count is currently an even number
thread2: increments count only when count is currently an odd number
Further Requirements:
-a thread prints the value of count AFTER the increment is performed.
-threads terminate when count reaches COUNT_DONE
-after threads terminate, main prints the final count (like cond1.c does).
-thread2 may test count to determine its parity, but thread1 may not.
-thread1 always increments count when it is signalled (unless it terminates)
-Your program output should look identical to this:
Counter value functionEven: 1
Counter value functionOdd: 2
Counter value functionEven: 3
Counter value functionOdd: 4
Counter value functionEven: 5
Counter value functionOdd: 6
Counter value functionEven: 7
Counter value functionOdd: 8
Counter value functionEven: 9
Counter value functionOdd: 10
Final count: 10
here is cond1.c for reference:
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
pthread_mutex_t count_mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t condition_var = PTHREAD_COND_INITIALIZER;
void *functionCount1();
void *functionCount2();
int count = 0;
#define COUNT_DONE 10
#define COUNT_HALT1 3
#define COUNT_HALT2 6
int main()
{
pthread_t thread1, thread2;
pthread_create( &thread1, NULL, &functionCount1, NULL);
pthread_create( &thread2, NULL, &functionCount2, NULL);
pthread_join( thread1, NULL);
pthread_join( thread2, NULL);
printf("Final count: %d ",count);
exit(0);
}
// Write numbers 1-3 and 8-10 as permitted by functionCount2()
void *functionCount1()
{
for(;;)
{
// Lock mutex and then wait for signal to relase mutex
pthread_mutex_lock( &count_mutex );
// Wait while functionCount2() operates on count
// mutex unlocked if condition varialbe in functionCount2() signaled.
pthread_cond_wait( &condition_var, &count_mutex );
count++;
printf("Counter value functionCount1: %d ",count);
pthread_mutex_unlock( &count_mutex );
if(count >= COUNT_DONE) return(NULL);
}
}
// Write numbers 4-7
void *functionCount2()
{
for(;;)
{
pthread_mutex_lock( &count_mutex );
if( count < COUNT_HALT1 || count > COUNT_HALT2 )
{
// Condition of if statement has been met.
// Signal to free waiting thread by freeing the mutex.
// Note: functionCount1() is now permitted to modify "count".
pthread_cond_signal( &condition_var );
}
else
{
count++;
printf("Counter value functionCount2: %d ",count);
}
pthread_mutex_unlock( &count_mutex );
if(count >= COUNT_DONE) return(NULL);
}
}
Explanation / Answer
#include <stdio.h>
#define MAX_SIZE 100 //Maximum size of the array
int main()
{
int arr[MAX_SIZE];
int i, N, even, odd;
/*
* Reads size and elements in array
*/
printf("Enter size of the array: ");
scanf("%d", &N);
printf("Enter %d elements in array: ", N);
for(i=0; i<N; i++)
{
scanf("%d", &arr[i]);
}
/* Assuming that there are 0 even and odd elements */
even = 0;
odd = 0;
for(i=0; i<N; i++)
{
/* If the current element of array is even then increment even count */
if(arr[i]%2 == 0)
{
even++;
}
else
{
odd++;
}
}
printf("Total even elements: %d ", even);
printf("Total odd elements: %d ", odd);
return 0;
}
#include <stdio.h>
#define MAX_SIZE 100 //Maximum size of the array
int main()
{
int arr[MAX_SIZE];
int i, N, even, odd;
/*
* Reads size and elements in array
*/
printf("Enter size of the array: ");
scanf("%d", &N);
printf("Enter %d elements in array: ", N);
for(i=0; i<N; i++)
{
scanf("%d", &arr[i]);
}
/* Assuming that there are 0 even and odd elements */
even = 0;
odd = 0;
for(i=0; i<N; i++)
{
/* If the current element of array is even then increment even count */
if(arr[i]%2 == 0)
{
even++;
}
else
{
odd++;
}
}
printf("Total even elements: %d ", even);
printf("Total odd elements: %d ", odd);
return 0;
}
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