Question

因此，我试图在两个数组之间进行简单的乘法，然后将每个乘法的结果相加，而归约让我很困惑，这是我的代码：

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问题是，如果我将“ Sum = Result [i]”更改为“ Sum + = Result [i]”，则会得到正确的结果。为什么会这样？难道不是对每个线程都创建并初始化了Sum的局部变量，然后在所有线程完成后，归约化将其加起来吗？

这是Sum + = Result [i]的结果：

#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

Answer 1

每个线程运行两次迭代，然后得出Sum的最终结果。因为您不是在每个迭代中都添加Sum，而是对其进行分配，所以对于该线程上次运行的Result[i]而言，最终结果将只是i。那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状态Sum，直到它们见面并把不同的Sum加在一起。

打开mp减少混乱

1 个答案: