Reputation: 105
I made this parallel code to share the iterations like first and last, fisrst+1 and last-1,... But I don't know how to improve the code in every one of the two parallel sections because I have an inner loop in the sections and I can't think of any way to simplify it, thanks.
This isn't about which values are stored in x or y, I use this sections design because the requisite is execute the iterations from 0 to N like: 0 N, 1 N-1, 2 N-2 but I would like to know if I can optimize the inner loops maintaining this model
int x = 0, y = 0,k,i,j,h;
#pragma omp parallel private(i, h) reduction(+:x, y)
{
#pragma omp sections
{
#pragma omp section
{
for (i=0; i<N/2; i++)
{
C[i] = 0;
for (j=0; j<N; j++)
{
C[i] += MAT[i][j] * B[j];
}
x += C[i];
}
}
#pragma omp section
{
for (h=N-1; h>=N/2; h--)
{
C[h] = 0;
for (k=0; k<N; k++)
{
C[h] += MAT[h][k] * B[k];
}
y += C[h];
}
}
}
}
x = x + y;
Upvotes: 0
Views: 246
Reputation: 61
The section construct is to distribute different tasks to different threads and each section block marks a different task so you will not be able to do that iterations in the order you want I answered you here:
Distribution of loop iterations between threads with a specific order
But I want to clarify that the requirement to use sections is that each block must be independent of the other blocks.
Upvotes: 1
Reputation: 5794
A section gets only one thread, so you can't make the loops parallel. How about
N
at the top level,x,y
?Although @Homer512 's solution looks correct to me too.
Upvotes: 0
Reputation: 13295
Using sections seems like the wrong approach. A pragma omp for
seems more appropriate. Also note that you forgot to declare j
private.
int x = 0, y = 0,k,i,j;
#pragma omp parallel private(i,j) reduction(+:x, y)
{
# pragma omp for nowait
for(i=0; i<N/2; i++) {
// local variable to make the life easier on the compiler
int ci = 0;
for(j=0; j<N; j++)
ci += MAT[i][j] * B[j];
x += ci;
C[i] = ci;
}
# pragma omp for nowait
for(i=N/2; i < N; i++) {
int ci = 0;
for(j=0; j<N; j++)
ci += MAT[i][j] * B[j];
y += ci;
C[i] = ci;
}
}
x = x + y;
Also, I'm not sure but if you just want x
as your final output, you can simplify the code even further:
int x=0, i, j;
#pragma omp parallel for reduction(+:x) private(i,j)
for(i=0; i < N; ++i)
for(j=0; j < N; ++j)
x += MAT[i][j] * B[j];
Upvotes: 2