如何同步cuda線程，當他們在同一個循環中，我們需要同步它們來執行只有有限的部分

Question

我寫了一個代碼，現在我想實現這個在cuda GPU上，但我是新同步所以請幫助我，這對我來說並不緊急。下面我將介紹代碼，並且我希望所有線程都執行LOOP1（我希望這部分利用cuda，剩下的部分（LOOP1中的其他部分）僅由一個單個線程。

do{ 
    point_set = master_Q[(*num_mas) - 1].q; 
    List* temp = point_set; 
    List* pa = point_set; 
    if(master_Q[num_mas[0] - 1].max) 
    max_level = (int) (ceilf(il2 * log(master_Q[num_mas[0] - 1].max))); 
    *num_mas = (*num_mas) - 1; 
    while(point_set){ 
    List* insert_ele = temp; 
    while(temp){ 
    insert_ele = temp; 
    if((insert_ele->dist[insert_ele->dist_index-1] <= pow(2, max_level-1)) || (top_level == max_level)){ 
    if(point_set == temp){ 
     point_set = temp->next; 
     pa = temp->next; 
    } 
    else{ 
     pa->next = temp->next; 
    } 
    temp = NULL; 
    List* new_point_set = point_set; 
    float maximum_dist = 0; 
    if(parent->p_index != insert_ele->point_index){ 
     List* tmp = new_point_set; 
     float *b = &(data[(insert_ele->point_index)*point_len]); 
**LOOP 1:**  while(tmp){ 
     float *c = &(data[(tmp->point_index)*point_len]); 
     float sum = 0.; 
     for(int j = 0; j < point_len; j+=2){ 
     float d1 = b[j] - c[j]; 
     float d2 = b[j+1] - c[j+1]; 
     d1 *= d1; 
     d2 *= d2; 
     sum = sum + d1 + d2; 
      } 
      tmp->dist[tmp->dist_index] = sqrt(sum); 
     if(maximum_dist < tmp->dist[tmp->dist_index]) 
      maximum_dist = tmp->dist[tmp->dist_index]; 
      tmp->dist_index = tmp->dist_index+1; 
      tmp = tmp->next; 
     } 
     max_distance = maximum_dist; 
    } 
    while(new_point_set || insert_ele){ 
     List* far, *par, *tmp, *tmp_new; 
     far = NULL; 
     tmp = new_point_set; 
     tmp_new = NULL; 
     float level_dist = pow(2, max_level-1); 
     float maxdist = 0, maxp = 0; 
     while(tmp){ 
     if(tmp->dist[(tmp->dist_index)-1] > level_dist){ 
     if(maxdist < tmp->dist[tmp->dist_index-1]) 
     maxdist = tmp->dist[tmp->dist_index-1]; 
     if(tmp == new_point_set){ 
     new_point_set = tmp->next; 
     par = tmp->next; 
     } 
     else{ 
     par->next = tmp->next; 
     } 
     if(far == NULL){ 
     far = tmp; 
     tmp_new = far; 
     } 
     else{ 
     tmp_new->next = tmp; 
     tmp_new = tmp; 
     } 
     if(parent->p_index != insert_ele->point_index) 
     tmp->dist_index = tmp->dist_index - 1; 
     tmp = tmp->next; 
     tmp_new->next = NULL; 
     } 
     else{ 
     par = tmp; 
     if(maxp < tmp->dist[(tmp->dist_index)-1]) 
     maxp = tmp->dist[(tmp->dist_index)-1]; 
     tmp = tmp->next; 
     } 

     } 
     if(0 == maxp){  
     tmp = new_point_set; 
     aloc_mem[*tree_index].p_index = insert_ele->point_index; 
     aloc_mem[*tree_index].no_child = 0; 
     aloc_mem[*tree_index].level = max_level--; 
     parent->children_index[parent->no_child++] = *tree_index; 
     parent = &(aloc_mem[*tree_index]); 
     tree_index[0] = tree_index[0]+1; 
     while(tmp){ 
     aloc_mem[*tree_index].p_index = tmp->point_index; 
     aloc_mem[(*tree_index)].no_child = 0; 
     aloc_mem[(*tree_index)].level = master_Q[(*cur_count_Q)-1].level; 
     parent->children_index[parent->no_child] = *tree_index; 
     parent->no_child = parent->no_child + 1; 
     (*tree_index)++; 
     tmp = tmp->next; 
     } 
     cur_count_Q[0] = cur_count_Q[0]-1; 
     new_point_set = NULL; 
     } 
     master_Q[*num_mas].q = far; 
     master_Q[*num_mas].parent = parent; 
     master_Q[*num_mas].valid = true; 
     master_Q[*num_mas].max = maxdist; 
     master_Q[*num_mas].level = max_level; 

      num_mas[0] = num_mas[0]+1; 
     if(0 != maxp){ 
     aloc_mem[*tree_index].p_index = insert_ele->point_index; 
     aloc_mem[*tree_index].no_child = 0; 
     aloc_mem[*tree_index].level = max_level; 
     parent->children_index[parent->no_child++] = *tree_index; 
     parent = &(aloc_mem[*tree_index]); 
     tree_index[0] = tree_index[0]+1; 
     if(maxp){ 
     int new_level = ((int) (ceilf(il2 * log(maxp)))) +1; 
     if (new_level < (max_level-1)) 
     max_level = new_level; 
     else 
     max_level--; 
     } 
     else 
     max_level--; 
     } 
     if(0 == maxp) 
     insert_ele = NULL; 
    } 

    } 
    else{ 
    if(NULL == temp->next){ 
     master_Q[*num_mas].q = point_set; 
     master_Q[*num_mas].parent = parent; 
     master_Q[*num_mas].valid = true; 
     master_Q[*num_mas].level = max_level; 
     num_mas[0] = num_mas[0]+1; 
    } 
    pa = temp; 
    temp = temp->next; 
    } 
    } 
    if((*num_mas) > 1){ 
    List *temp2 = master_Q[(*num_mas)-1].q; 
    while(temp2){ 
    List* temp3 = master_Q[(*num_mas)-2].q; 
    master_Q[(*num_mas)-2].q = temp2; 
    if((master_Q[(*num_mas)-1].parent)->p_index != (master_Q[(*num_mas)-2].parent)->p_index){ 
     temp2->dist_index = temp2->dist_index - 1; 
    } 
    temp2 = temp2->next; 
    master_Q[(*num_mas)-2].q->next = temp3; 
    } 
    num_mas[0] = num_mas[0]-1; 
    } 
    point_set = master_Q[(*num_mas)-1].q; 
    temp = point_set; 
    pa = point_set; 
    parent = master_Q[(*num_mas)-1].parent; 
    max_level = master_Q[(*num_mas)-1].level; 
    if(master_Q[(*num_mas)-1].max) 
    if(max_level > ((int) (ceilf(il2 * log(master_Q[(*num_mas)-1].max)))) +1) 
     max_level = ((int) (ceilf(il2 * log(master_Q[(*num_mas)-1].max)))) +1; 
    num_mas[0] = num_mas[0]-1; 
    } 

}while(*num_mas > 0); 

Answer 1

如果你想有一個單一的內核塊內的只有一個線程執行其代碼的部分，使用if語句只對單個threadIdx執行的代碼的一部分，然後屏障同步。也許你應該嘗試編寫一個內核併發布它供人們看看。