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Different behaviour between scalar and parallel code

I'm wondering why the following code produces different results in its scalar and parallel variants:

#define N 10
double P[N][N];
// zero the matrix just to be sure...
for (int i=0; i<N; i++)
    for(int j=0; j<N; j++)
        P[i][j]=0.0;


double xmin=-5.0,ymin=-5.0,xmax=5.0,ymax=5.0;
double x=xmin,y=ymin;
double step= abs(xmax-xmin)/(double)(N - 1 );
for (int i=0; i<N; i++)
{
    #pragma omp parallel for ordered schedule(dynamic)
    for ( int j=0; j<N; j++)
    {
        x = i*step+xmin;
        y = j*step+ymin;
        P[i][j]=x+y;
    }
}

This code produces not completely equal results in its two version (the scalar version has just the #pragma ... part commented out). What I've noticed is that a very small percentual of the elements of P[i][j] in the parallel version are different from those of the scalar version, but I'm wondering why...

Putting the #pragma on the outer loop as suggested is mess...completely wrong results.

P.S. g++-4.4, intel i7, linux

like image 852
linello Avatar asked Apr 18 '26 00:04

linello


1 Answers

Ah, now I can see the problem. Your comment on the last question didn't have enough context for me to see it. But now it's clear.

The problem is here:

    x = i*step+xmin;
    y = j*step+ymin;

x and y are declared outside the parallel region, so they are being shared among all the threads. (and thus a nasty race condition among all the threads...)

To fix it, make them local:

for ( int j=0; j<N; j++)
{
    double x = i*step+xmin;
    double y = j*step+ymin;
    P[i][j]=x+y;
}

With this fix, you should be able to put the #pragma on the outer loop instead of the inner loop.

like image 135
Mysticial Avatar answered Apr 19 '26 14:04

Mysticial



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