線程私有向量的openMP C++錯誤

Question

#include<iostream>
#include<vector>
#include<ctime>
#include<cmath>
#include<omp.h>
#include<array>
using namespace std;

double counter = 0;
vector<int> vec;

vector<double> reduction;
#pragma omp threadprivate(reduction)


vector<double> eet;
int fish;


void mario( int current_node, int current_depth) {
    int q = current_node % 4;
    #pragma omp simd
    for( int i = 3 ; i < current_node; i+=4){
        reduction[i]+=4;
        mario(i, current_depth);
        mario(i - 1,current_depth);
        mario(i - 2,current_depth);
        mario(i - 3,current_depth);
    }
    #pragma omp simd
    for(int x = 1; x <= q; x++){
        reduction[0]++;
        mario(current_node - x,current_depth);
    }
    #pragma omp task firstprivate(current_node,current_depth)
    {
        if(current_depth > 0){
                int new_depth = current_depth - 1;
                #pragma omp simd
                for(int i = current_node;i <= vec[current_node];i++){
                    reduction[i]++;
                    mario(i + 1,new_depth);
            }
        } 
    } 
}


int main() {
    omp_proc_bind_true;
    omp_set_dynamic(0);
    int nodes;
    int timesteps;
    int threadz;
    cout << "enter number of nodes" << endl;

    cin >> nodes;

    cout << "enter number of timesteps" << endl;

    cin >> timesteps;

    cout << "enter number threads" << endl;

    cin >> threadz;

    omp_set_num_threads(threadz);

    int mushroom = nodes - 2;
    fish = nodes - 1;

    vec.assign( nodes, mushroom );

    clock_t t = clock();

    vector<double> zed(mushroom + 1 , 0 );

    eet = zed;
    reduction = eet;
    #pragma omp parallel copyin(reduction)
    {

        #pragma omp single
        {
            mario(nodes - 1, timesteps - 1);
        }

        #pragma omp critical
        {
            #pragma omp simd
            for(int x = 0; x < zed.size();x++){
                eet[x] += reduction[x];
            }
        }
        #pragma omp barrier
    }
    for(int j = 0; j < eet.size(); j++){
            counter += eet[j];
    }
    t = clock() - t;
    double time_taken = ((double)t) / CLOCKS_PER_SEC;
    cout << "mario took " << fixed << counter << " steps" << endl;
    cout << "It took him " << time_taken << " seconds" << endl;
    return 0;
}

在我的代碼中，我聲明了向量縮減： vector<double> reduction;

然后我立即使用 threadprivate: #pragma omp threadprivate(reduction) ，但是當我編譯我的代碼時，我得到了錯誤。

錯誤：第一次使用后“reduction”聲明為“threadprivate”

13 | #pragma omp 線程私有（減少）

錯誤：對於“copyin”，“reduction”必須是“threadprivate”

82 | #pragma omp 並行復制（減少）

是什么導致了這些錯誤？ 我只是聲明了變量而沒有做任何其他事情。

錯誤是可以避免的嗎？

是否有任何其他方法可以減少遞歸函數的 openMP 任務

或解決問題的替代方法？

我嘗試編寫代碼的方式是否存在根本性錯誤？

我正在使用 gcc 9.3

，抱歉，如果我的問題不清楚，請提出改進建議。

Answer 1

僅適用於 gcc 的解決方案的一個組成部分，不幸的是，這只是一種解決方法，而不是適當的解決方案。

#include<iostream>
#include<vector>
#include<ctime>
#include<cmath>
#include<omp.h>
#include<array>
#include<thread>
using namespace std;

double counter = 0;
vector<int> vec;
thread_local vector<double> reduction;

vector<double> eet;
int fish;



void mario( int current_node, int current_depth) {
    int q = current_node % 4;
    #pragma omp simd
    for( int i = 3 ; i < current_node; i+=4){
        reduction[i]+=4;
        mario(i, current_depth);
        mario(i - 1,current_depth);
        mario(i - 2,current_depth);
        mario(i - 3,current_depth);
    }
    #pragma omp simd
    for(int x = 1; x <= q; x++){
        reduction[0]++;
        mario(current_node - x,current_depth);
    }
    #pragma omp task firstprivate(current_node,current_depth)
    {
        if(current_depth > 0){
                int new_depth = current_depth - 1;
                #pragma omp simd
                for(int i = current_node;i <= vec[current_node];i++){
                    reduction[i]++;
                    mario(i + 1,new_depth);
            }
        } 
    } 
}



int main() {
    omp_proc_bind_true;
    omp_set_dynamic(0);
    int nodes;
    int timesteps;
    int threadz;
    cout << "enter number of nodes" << endl;

    cin >> nodes;

    cout << "enter number of timesteps" << endl;

    cin >> timesteps;

    cout << "enter number threads" << endl;

    cin >> threadz;

    omp_set_num_threads(threadz);
    int mushroom = nodes - 2;
    fish = nodes - 1;
    vec.assign( nodes, mushroom );
    clock_t t = clock();
    vector<double> zed(mushroom + 1 , 0 );
    eet = zed;
    #pragma omp parallel
    {
        reduction = zed;
        #pragma omp barrier
        #pragma omp single
        {
            mario(nodes - 1, timesteps - 1);
        }
        #pragma omp critical
        {
            for(int x = 0; x < reduction.size();x++){
                eet[x] += reduction[x];
            }

        }
        #pragma omp barrier
    }
    for(int j = 0; j < eet.size(); j++){
            counter += eet[j];
    }
    t = clock() - t;
    double time_taken = ((double)t) / CLOCKS_PER_SEC;
    cout << "mario took " << fixed << counter << " steps" << endl;
    cout << "It took him " << time_taken << " seconds" << endl;
    return 0;
}

這是一種很糟糕的方法，顯然 gcc 的 c++ 實現使用 POSIX 線程，而 openMP 的 gcc 實現只是 POSIX 底層。 來源： 將 C++11 thread_local 與其他並行庫一起使用

因此，對於 gcc，如果以這種方式將變量發送到線程本地存儲，則它等效於 openMP 的“threadprivate”變量。

請其他人提出適當的解決方案。