打开 mp 减少令人困惑

Question

So I was trying to do simple multiplication between two arrays then add up the result of each multiplication and I am really confused by the reduction, here is my code:

#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;
}

The thing is, if i change the "Sum = Result[i]" to "Sum += Result[i]" I get the correct result. Why does this happen? Isn't a local variable of Sum made and initialized to each thread, then the reduction adds it all up when all the threads are done?

Here is the result with Sum += Result[i]:

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

And Here is the result with Sum = Result[i]:

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

Each thread is running through two iterations before coming to a final result for Sum . Because you are not adding to Sum each iteration, but rather assigning it, the final result will simply be Result[i] for whatever i was the last run on that thread. That is the value that is ultimately summed up with the results of all the other threads. You need Sum += Result[i] so that each thread keeps its own running Sum until they meet back up and add the different Sum s together.