open mp reduction令人困惑

Question

所以我试着在两个数组之间做简单的乘法，然后把每个乘法的结果加起来，我真的被减法搞糊涂了，这是我的代码：

#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;
}

问题是，如果我将"Sum = Resulti“更改为"Sum += Resulti”，我会得到正确的结果。这一切为什么要发生？Sum不是一个局部变量，并初始化到每个线程，然后在所有线程都完成时，缩减会将其相加吗？

以下是+=求和结果的结果

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

下面是Sum = Resulti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[3]= 448
Partial result by thread[3]= 1024
Partial result by thread[1]= 12
Partial result by thread[1]= 32
Final result= 1252

Answer 1

在得出Sum的最终结果之前，每个线程都会经过两次迭代。因为您不是在每次迭代中添加Sum，而是赋值它，所以对于上次在该线程上运行的任何i，最终结果都将是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]，以便每个线程保持其自己的运行Sum，直到它们遇到备份并将不同的Sum添加在一起。