通過openmp在C ++中並行化three_for_loop

Question

我有一個密碼。 在此，A，B，C，A1，B1，C1是3維矢量。 A，B，C相互獨立，而A1，B1，C1也相互獨立。 我想通過使用openmp進行並行計算。 但是，我用openmp運行它，出現“段錯誤”錯誤。您能幫我解決這個問題嗎？ 先感謝您。

#include <omp.h>
#include<math.h> 
#include<cmath> 
#include<vector>    
#include<iostream>

using namespace std;
int main ()
{

int NX=801;              // NUmber of grid in X direction
int NY=501;              
int NZ=401;   
float PI=3.14159265358979323846;
unsigned int i,j,k;
vector<vector<vector<float> > > A (NX,vector<vector<float> >(NY,vector <float>(NZ,0.0)));
vector<vector<vector<float> > > B (NX,vector<vector<float> >(NY,vector <float>(NZ,0.0)));
vector<vector<vector<float> > > C (NX,vector<vector<float> >(NY,vector <float>(NZ,0.0)));
vector<vector<vector<float> > > A1 (NX,vector<vector<float> >(NY,vector <float>(NZ,0.0)));
vector<vector<vector<float> > > B1 (NX,vector<vector<float> >(NY,vector <float>(NZ,0.0)));
vector<vector<vector<float> > > C1 (NX,vector<vector<float> >(NY,vector <float>(NZ,0.0)));

cout<<"start"<<endl;
#pragma omp parallel for private (j) shared(A,B,C,i,k,NX,NY,NZ) 
for (i=0;i<NX;i++)
    for (j=0;j<NY;j++)
        for (k=0;k<NZ;k++)
        {
            A[i][j][k]=sin(2.0*PI/float(NX*NY*NZ)*float(i*j*k));
            B[i][j][k]=cos(5.0*PI/float(NX*NY*NZ)*float(i*j*k));
            C[i][j][k]=sin(2.0*PI/float(NX*NY*NZ))*cos(5.0*PI/float(NX*NY*NZ)*float(i*j*k));
        }

#pragma omp parallel for private (j) shared(A1,B1,C1,A,B,C,i,k,NX,NY,NZ) 
for (i=1;i<NX-1;i++)
    for (j=1;j<NY-1;j++)
        for (k=1;k<NZ-1;k++)
        {
            A1[i][j][k]=C[i+1][j][k]*cos(5.0*PI/float(NX*NY*NZ)*float(i*j*k));
            B1[i][j][k]=A[i][j][k]+B[i][j][k]+C[i][j][k]*cos(5.0*PI/float(NX*NY*NZ)*float(i*j*k));
            C1[i][j][k]=16.0*A[i][j][k]*cos(5.0*PI/float(NX*NY*NZ)*float(i*j*k));
        }
cout<<"finish"<<endl;


return 0;
}

Answer 1

這段代碼很容易與OpenMP並行化。 但是，您在自己的嘗試中犯了一些錯誤，特別是嘗試聲明i和k shared而實際上它們應該是private 。 更好的是，不要預先聲明變量，而只是在for循環中聲明它們。 這樣，它們將自動具有正確的范圍，從而防止您混淆它。

這是它會給的：

#include <omp.h>
#include<math.h> 
#include<cmath> 
#include<vector>    
#include<iostream>

using namespace std;
int main ()
{

    int NX=801;              // NUmber of grid in X direction
    int NY=501;              
    int NZ=401;   
    float PI=3.14159265358979323846;
    vector<vector<vector<float> > > A (NX,vector<vector<float> >(NY,vector <float>(NZ,0.0)));
    vector<vector<vector<float> > > B (NX,vector<vector<float> >(NY,vector <float>(NZ,0.0)));
    vector<vector<vector<float> > > C (NX,vector<vector<float> >(NY,vector <float>(NZ,0.0)));
    vector<vector<vector<float> > > A1 (NX,vector<vector<float> >(NY,vector <float>(NZ,0.0)));
    vector<vector<vector<float> > > B1 (NX,vector<vector<float> >(NY,vector <float>(NZ,0.0)));
    vector<vector<vector<float> > > C1 (NX,vector<vector<float> >(NY,vector <float>(NZ,0.0)));

    cout<<"start"<<endl;
    #pragma omp parallel for
    for (int i=0;i<NX;i++)
        for (int j=0;j<NY;j++)
            for (int k=0;k<NZ;k++)
            {
                A[i][j][k]=sin(2.0*PI/float(NX*NY*NZ)*float(i*j*k));
                B[i][j][k]=cos(5.0*PI/float(NX*NY*NZ)*float(i*j*k));
                C[i][j][k]=sin(2.0*PI/float(NX*NY*NZ))*cos(5.0*PI/float(NX*NY*NZ)*float(i*j*k));
            }

    #pragma omp parallel for 
    for (int i=1;i<NX-1;i++)
        for (int j=1;j<NY-1;j++)
            for (int k=1;k<NZ-1;k++)
            {
                A1[i][j][k]=C[i+1][j][k]*cos(5.0*PI/float(NX*NY*NZ)*float(i*j*k));
                B1[i][j][k]=A[i][j][k]+B[i][j][k]+C[i][j][k]*cos(5.0*PI/float(NX*NY*NZ)*float(i*j*k));
                C1[i][j][k]=16.0*A[i][j][k]*cos(5.0*PI/float(NX*NY*NZ)*float(i*j*k));
            }
    cout<<"finish"<<endl;

    return 0;
}

現在，由於您要求並行化此代碼，所以我想您對性能感興趣。 因此，沒有什么可以阻止您實現這樣的一兩個非常基本的性能優化：

#include <omp.h>
#include<math.h> 
#include<cmath> 
#include<vector>    
#include<iostream>

using namespace std;
int main ()
{

    int NX=801;              // NUmber of grid in X direction
    int NY=501;              
    int NZ=401;   
    float PI=3.14159265358979323846;
    vector<vector<vector<float> > > A (NX,vector<vector<float> >(NY,vector <float>(NZ,0.0)));
    vector<vector<vector<float> > > B (NX,vector<vector<float> >(NY,vector <float>(NZ,0.0)));
    vector<vector<vector<float> > > C (NX,vector<vector<float> >(NY,vector <float>(NZ,0.0)));
    vector<vector<vector<float> > > A1 (NX,vector<vector<float> >(NY,vector <float>(NZ,0.0)));
    vector<vector<vector<float> > > B1 (NX,vector<vector<float> >(NY,vector <float>(NZ,0.0)));
    vector<vector<vector<float> > > C1 (NX,vector<vector<float> >(NY,vector <float>(NZ,0.0)));

    const float PIOverSize = PI/(NX*NY*NZ);
    const float sin2PIOverSize = sin(2.0f*PIOverSize);
    cout<<"start"<<endl;
    double tbeg = omp_get_wtime();
    #pragma omp parallel
    {
    #pragma omp for
    for (int i=0;i<NX;i++)
        for (int j=0;j<NY;j++)
        {
            float IJPIOverSize=i*j*PIOverSize;
            for (int k=0;k<NZ;k++)
            {
                A[i][j][k]=sin(2.0f*IJPIOverSize*k);
                B[i][j][k]=cos(5.0f*IJPIOverSize*k);
                C[i][j][k]=sin2PIOverSize*cos(5.0f*IJPIOverSize*k);
            }
         }
    #pragma omp for 
    for (int i=1;i<NX-1;i++)
        for (int j=1;j<NY-1;j++)
        {
            float IJPIOverSize=i*j*PIOverSize;
            for (int k=1;k<NZ-1;k++)
            {
                A1[i][j][k]=C[i+1][j][k]*cos(5.0f*IJPIOverSize*k);
                B1[i][j][k]=A[i][j][k]+B[i][j][k]+C[i][j][k]*cos(5.0f*IJPIOverSize*k);
                C1[i][j][k]=16.0f*A[i][j][k]*cos(5.0f*IJPIOverSize*k);
            }
        }
    }
    double time = omp_get_wtime() - tbeg;
    cout<<"finish in "<<time<<" seconds"<<endl;

    return 0;
}

這樣，您的代碼應該已經更快了。