对于if-else，increment，如何为OpenMP中任意数量的线程获得相同的并行输出？

Posted 2023-02-22

【中文标题】对于if-else，increment，如何为OpenMP中任意数量的线程获得相同的并行输出？

【英文标题】：How to get same output for parallelization for any number of threads in OpenMP for if-else, increment?

【发布时间】：2019-01-08 04:58:43

【问题描述】：

我想在没有任何 atomic 操作或 critical 部分的情况下并行化以下代码。

下面给出了我的代码，我已经尝试过但没有预期结果。对于任意数量的线程，我想要相同的输出，其中输出顺序可能不同，即使用 1 个线程的第 5 行可能出现在第 8 行，而使用 4 个线程。如果我没记错的话，这里的临界区是 k++ 区域。请建议我如何在 OpenMP 中做到这一点而不使用关键/原子类子句。

#include <iostream>
#include <vector>
#include "omp.h"
#include "cstdlib"
using namespace std;
int main(int argc, char *argv[])
    int t = atoi(argv[1]);
    vector<int> A0,3,5,7;
    vector<int> B1,2,3,0,3,0,3,0,1,2;
    int k = 0, j;
    for(int i = 0; i<A.size(); i++)
        k = A[i];
        #pragma omp parallel for schedule(static) num_threads(t)
        for(j = 0; j < A.size(); j++)
        if(j == B[k])
            printf("i(yes):%d # B[%d] = %d, thread = %d\n",i, k, B[k], omp_get_thread_num());
            if(k < B.size() && k < A[i+1])
                k++;
            
        else
            printf("i(no):%d # j = %d, k = %d, thread =  %d\n",i, j, k, omp_get_thread_num());
        
    

return 0;

` 当我使用 1 个线程运行时，我得到以下输出：

$ ./a.out 1
i(no):0 # j = 0, k = 0, thread =  0
i(yes):0 # B[0] = 1, thread = 0
i(yes):0 # B[1] = 2, thread = 0
i(yes):0 # B[2] = 3, thread = 0
i(yes):1 # B[3] = 0, thread = 0
i(no):1 # j = 1, k = 4, thread =  0
i(no):1 # j = 2, k = 4, thread =  0
i(yes):1 # B[4] = 3, thread = 0
i(yes):2 # B[5] = 0, thread = 0
i(no):2 # j = 1, k = 6, thread =  0
i(no):2 # j = 2, k = 6, thread =  0
i(yes):2 # B[6] = 3, thread = 0
i(yes):3 # B[7] = 0, thread = 0
i(no):3 # j = 1, k = 7, thread =  0
i(no):3 # j = 2, k = 7, thread =  0
i(no):3 # j = 3, k = 7, thread =  0

但是当我使用 2 个线程运行时，我得到以下不同步和预期的输出。

$ ./a.out 2
i(no):0 # j = 0, k = 0, thread =  0
i(yes):0 # B[0] = 1, thread = 0
i(no):0 # j = 2, k = 0, thread =  1
i(no):0 # j = 3, k = 1, thread =  1
i(no):1 # j = 2, k = 3, thread =  1
i(no):1 # j = 3, k = 3, thread =  1
i(yes):1 # B[3] = 0, thread = 0 
i(no):1 # j = 1, k = 4, thread =  0
i(no):2 # j = 2, k = 5, thread =  1
i(no):2 # j = 3, k = 5, thread =  1
i(yes):2 # B[5] = 0, thread = 0 
i(no):2 # j = 1, k = 6, thread =  0
i(yes):3 # B[7] = 0, thread = 0
i(no):3 # j = 2, k = 7, thread =  1
i(no):3 # j = 3, k = 7, thread =  1
i(no):3 # j = 1, k = 7, thread =  0

【参考方案1】：

你不能并行化你的循环，不管有没有原子。由于迭代j + 1 中k 的值取决于迭代j 中发生的情况，因此迭代必须按顺序运行。 （问问自己如果j == 1 迭代还没有增加k，那么j == 2 的迭代如何才能正常运行？）

您当前的代码在k 上存在竞争，当它被另一个线程更新时，它的值可能会在迭代中间发生变化。