如何制作依赖于-2的doacross循环

Question

我对OpenMP有意见。我必须做一个循环。例如：

    for (int i = 1; i < SIZE-2; i++) {
        for (int j = 2; j < SIZE-2; j++) {
            tab[i][j] = tab[i][j+2] + tab[i+2][j-2];
        }
    }

这里我依赖于j-2，j+2和i+2，我不知道如何解决这个依赖关系。

Answer 1

您可以尝试执行以下操作：

#pragma omp parallel for ordered(2)
for (int i = 1; i < SIZE-2; i++) {
    for (int j = 2; j < SIZE-2; j++) {
        #pragma omp ordered depend(sink:i,j+2) depend(sink:i+2,j-2) 
        tab[i][j] =  tab[i][j+2] + tab[i+2][j-2];
        #pragma omp ordered depend(source)
    }

}

Answer 2

我得到了一个基于answer by dreamcrash的工作解决方案

#pragma omp parallel for ordered(2)
for(int i=1; i<N-2; i++){
    for(int j=1; j<N-2; j++){
        #pragma omp ordered depend(sink:i,j-2) depend(sink:i-2,j+1)
        a[i][j] = a[i][j+2] + a[i+2][j-1];
        #pragma omp ordered depend(source)   
    }
}