Question

//In this assignment, we use multiple threads to calculate the sum
// 1*1 + 2*2 + 3*3 + 4*4 + ... + n*n
// Note we should know from CSE2500 that this sum is
// n*(n+1)*(2*n+1)/6
// We a n value, we will create 2*n threads to do the calculation so that
// we can have a race condition.
// Before you change the code, read the code and run the code and see what
// happens. Do we already get the result sum?
// Read the code carefully and make changes to the code.
// Use mutex to ensure we always get the right sum.

#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <pthread.h>
#include <unistd.h>
#define MAX 1000

int n;
unsigned long sum = 0;
pthread_mutex_t mutex;

void* calculate(void* threadarg)
{

//fill in one line of code blew

if(n > 0)
{
sum += n*n;
//do not delete the following line of code
//this line helps us to see wrong results
sleep(0);
n--;
}
//fill in one line of code below
  
pthread_exit(NULL);
}

int main(int argc, char *argv[])
{

if(argc!=2)
{
printf("Usage: %s n (1 - 500)\n", argv[0]);
return -1;
}

n = atoi(argv[1]);
assert(n >= 1 && n <= MAX/2);

//save n to k for later use
int k = n;
int m = 2*n;

//fill in one line of code below

pthread_t threads[MAX];
int rc, t;
for( t=0; t<m; t++ ) {
//Since we do not need to pass any arguments to the thread, we pass a NULL
rc = pthread_create(&threads[t], NULL, calculate, NULL);
if (rc) {
printf("ERROR; return code from pthread_create() is %d\n", rc);
exit(-1);
}
}

for( t=0; t<m; t++ )
{
rc = pthread_join( threads[t], NULL );
if( rc ){
printf("ERROR; return code from pthread_join() is %d\n", rc);
exit(-1);
}
}
//fill in one line of code below

unsigned long correct_sum = k*(k+1)*(2*k+1)/6;
printf("thread sum : %ld correct sum : %ld\n", sum, correct_sum);
return 0;
}

Answer 1

//In this assignment, we use multiple threads to calculate the sum
// 1*1 + 2*2 + 3*3 + 4*4 + ... + n*n
// Note we should know from CSE2500 that this sum is
// n*(n+1)*(2*n+1)/6
// We a n value, we will create 2*n threads to do the calculation so that
// we can have a race condition.
// Before you change the code, read the code and run the code and see what
// happens. Do we already get the result sum?
// Read the code carefully and make changes to the code.
// Use mutex to ensure we always get the right sum.

#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <pthread.h>
#include <unistd.h>
#define MAX 1000

int n;
unsigned long sum = 0;
pthread_mutex_t mutex;

void* calculate(void* threadarg)
{

//fill in one line of code blew
pthread_mutex_lock(&mutex);
if(n > 0)
{
sum += n*n;
//do not delete the following line of code
//this line helps us to see wrong results
sleep(0);
n--;
}
//fill in one line of code below
pthread_mutex_unlock(&mutex);
pthread_exit(NULL);
}

int main(int argc, char *argv[])
{

if(argc!=2)
{
printf("Usage: %s n (1 - 500)\n", argv[0]);
return -1;
}

n = atoi(argv[1]);
assert(n >= 1 && n <= MAX/2);

//save n to k for later use
int k = n;
int m = 2*n;

//fill in one line of code below
if (pthread_mutex_init(&mutex, NULL) != 0)
{
printf("\n mutex init has failed\n");
return 1;
}

pthread_t threads[MAX];
int rc, t;
for( t=0; t<m; t++ ) {
//Since we do not need to pass any arguments to the thread, we pass a NULL
rc = pthread_create(&threads[t], NULL, calculate, NULL);
if (rc) {
printf("ERROR; return code from pthread_create() is %d\n", rc);
exit(-1);
}
}

for( t=0; t<m; t++ )
{
rc = pthread_join( threads[t], NULL );
if( rc ){
printf("ERROR; return code from pthread_join() is %d\n", rc);
exit(-1);
}
}
//fill in one line of code below
pthread_mutex_destroy(&mutex);
unsigned long correct_sum = k*(k+1)*(2*k+1)/6;
printf("thread sum : %ld correct sum : %ld\n", sum, correct_sum);
return 0;
}