C 开放式mp还原

所以我试着在两个数组之间进行简单的乘法，然后将每次乘法的结果相加，我真的被简化弄糊涂了，下面是我的代码：

#include <omp.h>
#include <stdio.h>
#define SizeOfVector 8
#define NumberOfThreads 4
int main(){
    const int X[SizeOfVector] = {0,2,3,4,5,6,7,8};
    const int Y[SizeOfVector] = {1,2,4,8,16,32,64,128};
    int Result[SizeOfVector] = {0};
    int Sum = 0;
    unsigned short id;

    omp_set_num_threads(NumberOfThreads);

    #pragma omp parallel private(id)
    {
        id = omp_get_thread_num();

        #pragma omp for reduction(+:Sum)
        for(unsigned short i = 0; i < SizeOfVector; i++)
        {
            Result[i] = X[i] * Y[i];
            Sum = Result[i];    //Problem Here
            printf("Partial result by thread[%d]= %d\n", id, Result[i]);
        }
    }
    printf("Final result= %d\n", Sum);
    return 0;
}

这是Sum=result[i]的结果：

Partial result by thread[2]= 80
Partial result by thread[2]= 192
Partial result by thread[0]= 0
Partial result by thread[0]= 4
Partial result by thread[1]= 12
Partial result by thread[1]= 32
Partial result by thread[3]= 448
Partial result by thread[3]= 1024
Final result= 1792

Partial result by thread[2]= 80
Partial result by thread[2]= 192
Partial result by thread[0]= 0
Partial result by thread[0]= 4
Partial result by thread[3]= 448
Partial result by thread[3]= 1024
Partial result by thread[1]= 12
Partial result by thread[1]= 32
Final result= 1252

---

每个线程运行两次迭代，然后得出

Sum

的最终结果。因为您不是对每个迭代进行相加，而是对其赋值，所以对于该线程上最后一次运行的

i

，最终结果将只是

result[i]

。这是最终与所有其他线程的结果相加的值。您需要

Sum+=Result[i]

以便每个线程保持自己的运行

Sum

，直到它们相遇备份并将不同的

Sum

添加到一起。

reduce子句仅创建变量的初始化私有副本，并在构造结束时将所有私有变量聚合为单个结果。它不会修改在构造内部使用私有变量的方式。