BSP implementation Sieve Of Erastothenes doesn't print C

Question

I'm working on implementing a parallel algorithm of the Sieve Of Erastothenes in C using BSP.

My code compiles and executes but doesn't print the primes. When I do ./bspsieve 2 1000 for example the only output I get is "It took : 0.000371 seconds for proc 0 out of 2." while it should be printing all the found primes! Oddly enough the algorithm does work it seems like. If I use a larger upperbound in the above example, it takes longer. When I assign more processors, it takes less time. So probably a stupid mistake somewhere, but I have serious problems with C and am working on a remote computer so unsure of that tools I've got...

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <mcbsp.h>


/*
Note: To compile, this file has to be in the same folder as mcbsp.h and you need the 2 following commands:
gcc -Iinclude/ -pthread -c -o bspsieve.o bspsieve.c
gcc -o bspsieve bspsieve.o lib/libmcbsp1.1.0.a -lpthread -lrt

*/

  int procs;
  int upperbound;
  int *primes;

//SPMD function
void bspSieve(){
    bsp_begin(procs);

    int p = bsp_nprocs(); // p = number of procs obtained 
    int s = bsp_pid();  // s = proc number

    float blocksize;    // block size to be used, note last proc has a different size!
    if( s != p-1){
        blocksize = ceil(upperbound/p); 
    } else {
        blocksize = upperbound - (p-1)*ceil(upperbound/p);
    }

    // Initialize start time and end time, set start time to now.   
    double start_time,end_time;
    start_time = bsp_time();

    // Create vector that has block of candidates
    int *blockvector;
    blockvector = (int *)malloc(blocksize*sizeof(int));
    int q;
    for(q = 0; q<blocksize; q++){
    //List contains the integers from s*blocksize till blocksize + s*blocksize
      blockvector[q] = q + s*blocksize;
    } 

    //We neglect the first 2 'primes' in processor 0.
     if(s == 0){
        blockvector[0] = 0;
        blockvector[1] = 0;
     }

    // We are using the block distribution. We assume that n is large enough  to
    // assure that n/p is larger than sqrt(n). This means that we will always find the
    // sieving prime in the first block, and so have to broadcast from the first 
    // processor to the others.
     long sieving_prime;
     int i;
     bsp_push_reg( &sieving_prime,sizeof(long) );
     bsp_sync();

     for(i = 2; i * i < upperbound; i++) {
       //Part 1: if first processor, get the newest sieving prime, broadcast. Search for newest prime starting from i.
        if(s == 0){
       int findPrimeNb;
       for(findPrimeNb = i; findPrimeNb < blocksize; findPrimeNb++) {
          if( blockvector[findPrimeNb] != 0) {
                 sieving_prime  = blockvector[findPrimeNb];
                 //broadcast
         int procNb;
         for(procNb = 0; procNb < p; ++procNb){
                     bsp_put(procNb, &sieving_prime,&sieving_prime,0,sizeof(long));
        }
             break;
          }
           }
    }
        bsp_sync();

    //Part 2: Sieve using the sieving prime
    int sievingNb;
    for(sievingNb = 0; sievingNb < blocksize; sievingNb++){
       //check if element is multiple of sieving prime, if so, pcross out (put to zero)
       if( blockvector[sievingNb] % sieving_prime == 0){
          blockvector[sievingNb] = 0;
           }
    }

     }

     //part 3: get local primes to central area
     int transferNb;
     long transferPrime;
     for(transferNb = 0; transferNb < blocksize; transferNb++){
        transferPrime = blockvector[transferNb];
        primes[transferPrime] = transferPrime;
     }

     // take the end time.
     end_time = bsp_time();

   //Print amount of taken time, only processor 0 has to do this.
   if( s == 0 ){
      printf("It took : %.6lf seconds for proc %d out of %d. \n", end_time-start_time, bsp_pid(), bsp_nprocs());
      fflush(stdout);
   }
   bsp_pop_reg(&sieving_prime);
   bsp_end();
}



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

    if(argc != 3) {
    printf( "Usage: %s <proc count> <upper bound> <n", argv[ 0 ] );
    exit(1);
    }
    //retrieve parameters
    procs = atoi( argv[ 1 ] );
    upperbound = atoi( argv[ 2 ] );

    primes = (int *)malloc(upperbound*sizeof(int));

    // init and call parallel part
    bsp_init(bspSieve, argc, argv); 
    bspSieve();


   //Print all non zeros of candidates, these are the primes.
   // Primes only go to p*p <= n
   int i;
   for(i = 0; i*i <= upperbound; i++) {
        if(primes[i] > 0) {
        printf("%d, ",primes[i]);
    }
    }
    return 0;
}

Answer 1

printf doesn't automatically include a trailing newline, and it doesn't generally flush the output buffer until it outputs a newline; so probably you just need to add a

printf("\n");

at the end of your program, just before your return 0;.

Alternatively, or additionally, if you want to see the output as-you-go (if BSP allows that), you can add a

fflush(stdout);

right after your printf("%d, ",primes[i]);, to explicitly flush the output buffer.