Ατζέντα εργαστηρίου Παράλληλου Προγραμματισμού 11/5/2021

README.md

Ατζέντα εργαστηρίου Παράλληλου Προγραμματισμού 11/5/2021

Στο gist αυτό θα αναρτηθούν λύσεις και υποδείξεις κατά τη διεξαγωγή του εργαστηρίου.

S10.c

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

#include <omp.h>

#define N 10
# define R 2

// Simple demo of parallel for

// compile with:
// gcc -O2 -Wall -fopenmp omp-separate-for-example.c -o omp-separate-for-example


int main() {

  #pragma omp parallel
  {
    for (int j=0;j<R;j++) {
      
      #pragma omp for nowait
      for (int i=0;i<N;i++) {
    
        printf("Thread %d working on element %d pass=%d\n",omp_get_thread_num(),i,j);
        
      }
      
    }
  } 

  return 0;
}

S20.c

// Simple sum-reduction example - no special OpenMP reduction is used.
// Compile with: gcc -O2 -Wall -fopenmp sum-reduction-omp.c -o sum-reduction-omp -DN=10000000

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

#include <omp.h>

int main() {
double *a;

  // allocate array
  if ((a = (double *)malloc(N*sizeof(double)))==NULL) {
    printf("Allocation failed!\n");
    exit(1);
  }	  

  // init array to 1..N
  for (int i=0;i<N;i++) {
    a[i] = i+1;
  }
  
  // reduce array to sum
  double sum = 0;
  
  #pragma omp parallel shared(sum)
  {
     double mypart = 0.0;	// private by default
  
     #pragma omp for
     for (int i=0;i<N;i++) {
       mypart += a[i];
     }	// implicit barrier here
     
     #pragma omp critical
     {
       sum += mypart;
     }
  }

  // check result
  double result = ((double)N*(N+1))/2;  
  if (sum!=result) {
    printf("Reduction error!\n");
  }

  // free array
  free(a);
  
  return 0;
}

S30.c

// Simple sum-reduction example - OpenMP reduction is used.
// Compile with: gcc -O2 -Wall -fopenmp sum-reduction-omp2.c -o sum-reduction-omp2 -DN=10000000

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

#include <omp.h>

int main() {
double *a;

  // allocate array
  if ((a = (double *)malloc(N*sizeof(double)))==NULL) {
    printf("Allocation failed!\n");
    exit(1);
  }	  

  // init array to 1..N
  for (int i=0;i<N;i++) {
    a[i] = i+1;
  }
  
  // reduce array to sum
  double sum = 0;
  
  #pragma omp parallel for reduction(+:sum)
  for (int i=0;i<N;i++) {
       sum += a[i];
  }

  // check result
  double result = ((double)N*(N+1))/2;  
  if (sum!=result) {
    printf("Reduction error!\n");
  }

  // free array
  free(a);
  
  return 0;
}