pthreads: using multiple threads to calculate successive prime numbers - c

Question

I am trying to get the example code to work such that multiple threads will calculate the sum of successive prime numbers (note that the original author's algorithm for successive prime calculation is very inefficient). So far, running unit tests shows that the output is inconsistent, i.e. it will change slightly each time I run the program. I will post the modified source code in C, along with output for debugging purposes.
Source:
/************************************************************************
* Code listing from "Advanced Linux Programming," by CodeSourcery LLC *
* Copyright(C) 2001 by New Riders Publishing *
* See COPYRIGHT for license information. *
***********************************************************************/
/*
* Modified By : Dylan Gleason
* Class : CST 352 - Operating Systems
* Date : 10/18/2012
*/
#include <pthread.h>
#include <stdlib.h>
#include <stdio.h>
#define DEBUG 0 /* Set to 1 to enable debug statements */
/* global variables to be accessed by each thread */
int current_sum = 2;
int primes_to_compute = 0;
/* create mutex for ensuring serial access to global data */
int thread_flag;
pthread_cond_t cond;
pthread_mutex_t lock;
/* print the thread info for debugging purposes */
void print_thread_info()
{
printf("Current thread ID : %u\n",(unsigned int*)pthread_self());
printf("Current sum of primes : %d\n", current_sum);
printf("Current prime to compute : %d\n\n", primes_to_compute);
}
/* initialize the mutex and return an integer value to determine if
initialization failed or not */
int initialize_mutex()
{
int success = 1;
if(pthread_mutex_init(&lock, NULL) == 0 &&
pthread_cond_init(&cond, NULL) == 0)
success = 0;
thread_flag = 0;
return success;
}
/* set the value of the wait thread flag to the value which the client
passes */
void set_thread_flag(int is_waiting)
{
pthread_mutex_lock(&lock); /* lock mutex */
/* set the wait flag value, and then signal in case the prime
function is blocked, waiting for flag to become set. However,
prime function can't actually check flag until the mutex is
unlocked */
thread_flag = is_waiting;
pthread_cond_signal(&cond);
pthread_mutex_unlock(&lock); /* unlock mutex */
}
void in_wait()
{
while(!thread_flag)
pthread_cond_wait(&cond, &lock);
}
/* Compute successive prime numbers(very inefficiently). Return the
Nth prime number, where N is the value pointed to by *ARG. */
void* compute_prime(void* arg)
{
while(1)
{
pthread_mutex_lock(&lock);
in_wait();
pthread_mutex_unlock(&lock);
int sum;
int factor;
int is_prime = 1;
set_thread_flag(0);
pthread_mutex_lock(&lock);
sum = current_sum;
if(DEBUG)
{
printf("First lock\n");
print_thread_info();
}
pthread_mutex_unlock(&lock);
set_thread_flag(1); /* tell next thread to go! */
/* wait until ready-flag is released from current thread */
pthread_mutex_lock(&lock);
in_wait();
pthread_mutex_unlock(&lock);
/* Test primality by successive division. */
for(factor = 2; factor < sum; ++factor)
{
if(sum % factor == 0)
{
is_prime = 0;
break;
}
}
/* Is this the prime number we're looking for? */
if(is_prime)
{
int number;
set_thread_flag(0);
pthread_mutex_lock(&lock);
/* only decrement primes_to_compute if is greater than zero! */
if(primes_to_compute > 0)
{
--primes_to_compute;
}
if(DEBUG)
{
printf("Second lock\n");
print_thread_info();
}
number = primes_to_compute;
pthread_mutex_unlock(&lock);
set_thread_flag(1);
pthread_mutex_lock(&lock);
in_wait();
pthread_mutex_unlock(&lock);
if(number == 0)
{
set_thread_flag(0);
pthread_mutex_lock(&lock);
void* sum =(void*) current_sum;
if(DEBUG)
{
printf("Third lock\n");
print_thread_info();
}
pthread_mutex_unlock(&lock);
set_thread_flag(1);
return sum;
}
}
set_thread_flag(0);
pthread_mutex_lock(&lock);
++current_sum;
if(DEBUG)
{
printf("Fourth lock\n");
print_thread_info();
}
pthread_mutex_unlock(&lock);
set_thread_flag(1);
}
return NULL;
}
int main(int argc, char* argv[])
{
int prime;
pthread_t tid[5];
/* Check command-line argument count */
if(argc != 2)
{
printf("Error: wrong number of command-line arguments\n");
printf("Usage: %s <integer>\n", argv[0]);
exit(1);
}
/* Check to see if mutex initialized correctly */
if(initialize_mutex() != 0)
{
printf("Mutex initialization failed.\n");
exit(1);
}
primes_to_compute = atoi(argv[1]);
printf("Successive primes to be computed: %d\n\n", primes_to_compute);
/* Execute five different threads to calculate the prime summation */
int t = 0;
set_thread_flag(1);
for(t; t < 5; ++t)
pthread_create(&tid[t], NULL, &compute_prime, NULL);
/* Wait for the prime number thread to complete, then get result. */
t = 0;
for(t; t < 5; ++t)
pthread_join(tid[t],(void*) &prime);
/* Print the largest prime it computed. */
printf("The %dth prime number is %d.\n", atoi(argv[1]), prime);
return 0;
}
Unit test (executing program five times):
Test successive primes up to 100:
Successive primes to be computed: 100
The 100th prime number is 547.
Successive primes to be computed: 100
The 100th prime number is 521.
Successive primes to be computed: 100
The 100th prime number is 523.
Successive primes to be computed: 100
The 100th prime number is 499.
Successive primes to be computed: 100
The 100th prime number is 541.
Note that the output of non-threaded version if the number for successive primes to calculate is 100, the result will always be 541. Clearly I am not able to grok the correct usage of the mutexes above - if someone has more experience in this area I would be very grateful! Also, please note that I am not concerned with the efficiency/correctness of the actual prime number algorithm, but rather the algorithm for making sure that the threads execute properly with consistent results.

Answer 1

Ok, looking at your program I think I see what's going on.
You have a race condition, and a pretty bad one. Which number you're on is determined by the current_sum variable. You access it at the beginning of each loop, but don't increment it until the end of the loop. You need to set and then increment it at the same time within the same mutex lock, otherwise two different threads will be able to pull the same value, and if they pull the same prime value then that prime will be counted twice.
Hope this helps.