CODEWITHSUNDEEP
c++multithreadingmultiprocessingopenmp
Gilad
Gilad

Reputation: 6595

openMP exercise omp_bug2.c

this is an exercise from the OpenMP website: https://computing.llnl.gov/tutorials/openMP/exercise.html


#include "stdafx.h"
#include <omp.h>
#include <stdio.h>
#include <stdlib.h>

int _tmain(int argc, _TCHAR* argv[])
{
int nthreads, i, tid;
float total;

/*** Spawn parallel region ***/
#pragma omp parallel private(i, tid) // i changed this line 
  {
  /* Obtain thread number */
  tid = omp_get_thread_num();
  /* Only master thread does this */
  if (tid == 0) {
    nthreads = omp_get_num_threads();
    printf("Number of threads = %d\n", nthreads);
    }
  printf("Thread %d is starting...\n",tid);

  #pragma omp barrier

  /* do some work */
  total = 0.0;
  #pragma omp for schedule(dynamic,10) 
  for (i=0; i<1000000; i++)
     total = total + i*1.0;

  printf ("Thread %d is done! Total= %e\n",tid,total);

  } 
}

the output for this is 

Number of threads = 4
Thread 0 is starting...
Thread 3 is starting...
Thread 2 is starting...
Thread 1 is starting...
Thread 0 is done! Total= 0.000000e+000
Thread 3 is done! Total= 0.000000e+000
Thread 2 is done! Total= 0.000000e+000
Thread 1 is done! Total= 0.000000e+000

which means we have a problem with the variable "total"

this is the help on the site enter image description here

Here is my Solution: do you think this is the correct way to do it?

#include "stdafx.h"
#include <omp.h>
#include <stdio.h>
#include <stdlib.h>

int _tmain(int argc, _TCHAR* argv[])
{
int nthreads, i, tid;
float total;

/*** Spawn parallel region ***/
#pragma omp parallel private(total,tid)
  {
  /* Obtain thread number */
  tid = omp_get_thread_num();
  total= 0.0;
  /* Only master thread does this */
  if (tid == 0) {
    nthreads = omp_get_num_threads();
    printf("Number of threads = %d\n", nthreads);
    }
  printf("Thread %d is starting...\n",tid);



  #pragma omp parallel for schedule(static,10)\
  private(i)\
  reduction(+:total)
  for (i=0; i<1000000; i++) 
     total = total + i*1.0;

  printf ("Thread %d is done! Total= %e\n",tid,total);

  } /*** End of parallel region ***/
}

Here is my new output:

Number of threads = 4
Thread 0 is starting...
Thread 1 is starting...
Thread 0 is done! Total= 4.999404e+011
Thread 2 is starting...
Thread 1 is done! Total= 4.999404e+011
Thread 2 is done! Total= 4.999404e+011
Thread 3 is starting...
Thread 3 is done! Total= 4.999404e+011

Upvotes: 0

Views: 299

Answers (1)

Voo
Voo

Reputation: 30216

Yes you certainly want total to be a thread-private variable. One thing you presumably would do in a real example is to reduce the thread-private totals to a single global total at the end (and only let one thread print the result then). One way to do that is a simple 

#pragma omp atomic
global_total += total

at the end (there are better ways though using reductions).

PS: Loop counters for omp for are by default private, so you actually don't have to explicitly specify that.

Upvotes: 1

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