我对C中的OpenMP是个新手。我用它来在函数中并行我的for循环,但事实证明,与单线程的情况相比,它大大减慢了我的for循环。例如,对于每个点(光环),For循环可以在10s左右完成,但使用OpenMP需要几分钟。
在这个函数中,我试图计算每个点(光晕)的许多壳的密度,通过计数壳内的粒子,然后将它们写到一个数组中。我要计算的粒子有512^3个,大约有200个点(光晕)。我想拆分不同线程的点(光环),以使其更快。
#define ArrayAccess2D_n2(a, n1, n2, i1, i2) (a)[ i2+n2*i1 ]
void halo_shell_rho(float boxsize, float *halo_pos, float *halo_R, int halo_number,\
int halo_start, int halo_end, float *par_pos, long long par_number,\
int shell_bins, float rmax_fac, float *out_shell_den){
float temp;
long long iter_sfs=0, iter_sfc=0, iter_ufs=0, iter_ufc=0;
int dim=3;
float par_posx, par_posy, par_posz, dist;
float halo_posx, halo_posy, halo_posz, halo_rad;
int i=0, ini_j=0, vol_j=0;
int a=0, b=0;
long long k=0;
#pragma omp parallel for private(i, ini_j, vol_j, a, b, k)
for(i=halo_start; i<=halo_end; i++){
printf("halo %d\n", i);
float count[shell_bins];
float volume[shell_bins];
for(ini_j=0; ini_j<shell_bins; ini_j++){
count[ini_j] = 0;
volume[ini_j] = 0; }
halo_posx = ArrayAccess2D_n2(halo_pos, dim, halo_number, 0, i);
halo_posy = ArrayAccess2D_n2(halo_pos, dim, halo_number, 1, i);
halo_posz = ArrayAccess2D_n2(halo_pos, dim, halo_number, 2, i);
halo_rad = halo_R[i];
for(vol_j=0; vol_j<shell_bins; vol_j++){
volume[vol_j] = shell_volume((vol_j+1)*halo_rad*rmax_fac/(shell_bins*1000), vol_j*halo_rad*rmax_fac/(shell_bins*1000)); }
for(k=0; k<par_number; k++){
par_posx = ArrayAccess2D_n2(par_pos, par_number, dim, k, 0);
par_posy = ArrayAccess2D_n2(par_pos, par_number, dim, k, 1);
par_posz = ArrayAccess2D_n2(par_pos, par_number, dim, k, 2);
dist = pb_distance(boxsize*1000, halo_posx, halo_posy, halo_posz, par_posx, par_posy, par_posz); //1000 for boxsize in Mpc
if(dist <= 2*rmax_fac*halo_rad){
for(a=0; a<shell_bins; a++){
if((dist <= halo_rad*(a+1)*rmax_fac/shell_bins) && (dist >= halo_rad*a*rmax_fac/shell_bins)){
count[a] += 1; }
}
}
}
for(b=0; b<shell_bins; b++){
out_shell_den[(i-halo_start+0*(1+halo_end-halo_start))*shell_bins+b] = count[b]/volume[b];
//out_shell_den has shape (2, halo_number, shell_bins), 0 for edge, 1 for density
out_shell_den[(i-halo_start+1*(1+halo_end-halo_start))*shell_bins+b] = (2*b+1)*rmax_fac/(shell_bins*2);
}
}}
有人能帮我解决这个问题吗?我知道这是一个经常被问到的问题,但我没有从其他帖子中找到任何解决方案。如果有帮助,我在一个有32个线程的集群上运行它。
谢谢!
发布于 2017-08-05 09:49:51
感谢@DavidSchwartz和@tim18。
像halo_rad和par_posx这样的变量是在并行之前声明的,这意味着它们被隐式假设为公共的。所以它变慢了,因为所有的线程都在争夺它们的使用权。解决这个问题的一种方法是将所有变量添加到private()中。但我认为更好的方法是在并行中声明变量,如下所示:
void halo_shell_rho(float boxsize, float *halo_pos, float *halo_R, int halo_number, int halo_start, int halo_end, float *par_pos, long long par_number, int shell_bins, float rmax_fac, float *out_shell_den){
int dim=3;
int i=0, ini_j=0, vol_j=0, a=0, b=0;
long long k=0;
#pragma omp parallel for private(i, ini_j, vol_j, a, b, k)
for(i=halo_start; i<=halo_end; i++){
printf("halo %d\n", i);
float halo_posx, halo_posy, halo_posz, halo_rad;
float count[shell_bins];
float volume[shell_bins];
for(ini_j=0; ini_j<shell_bins; ini_j++){
count[ini_j] = 0;
volume[ini_j] = 0; }
halo_posx = ArrayAccess2D_n2(halo_pos, dim, halo_number, 0, i);
halo_posy = ArrayAccess2D_n2(halo_pos, dim, halo_number, 1, i);
halo_posz = ArrayAccess2D_n2(halo_pos, dim, halo_number, 2, i);
halo_rad = halo_R[i];
for(vol_j=0; vol_j<shell_bins; vol_j++){
volume[vol_j] = shell_volume((vol_j+1)*halo_rad*rmax_fac/(shell_bins*1000), vol_j*halo_rad*rmax_fac/(shell_bins*1000)); }
for(k=0; k<par_number; k++){
float par_posx, par_posy, par_posz, dist;
par_posx = ArrayAccess2D_n2(par_pos, par_number, dim, k, 0);
par_posy = ArrayAccess2D_n2(par_pos, par_number, dim, k, 1);
par_posz = ArrayAccess2D_n2(par_pos, par_number, dim, k, 2);
dist = pb_distance(boxsize*1000, halo_posx, halo_posy, halo_posz, par_posx, par_posy, par_posz); //1000 for boxsize in Mpc
if(dist <= 2*rmax_fac*halo_rad){
for(a=0; a<shell_bins; a++){
if((dist <= halo_rad*(a+1)*rmax_fac/shell_bins) && (dist >= halo_rad*a*rmax_fac/shell_bins)){
count[a] += 1; }
}
}
}
for(b=0; b<shell_bins; b++){
out_shell_den[(i-halo_start+0*(1+halo_end-halo_start))*shell_bins+b] = count[b]/volume[b]; //out_shell_den has shape (2, halo_number, shell_bins), 0 for edge, 1 for density
out_shell_den[(i-halo_start+1*(1+halo_end-halo_start))*shell_bins+b] = (2*b+1)*rmax_fac/(shell_bins*2);
}
}
}https://stackoverflow.com/questions/45516488
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