Question

此代码按预期工作：

#include <iostream>
#include <cmath>
#include <omp.h>

//https://stackoverflow.com/questions/37970024/jacobi-relaxation-in-mpi
#define max(a, b) (a)>(b)?(a):(b)

const int m = 2001;
const int n = 1500;
const int p = 4;

double v[m + 2][m + 2];
double x[m + 2];
double y[m + 2];
double _new[m + 2][m + 2];
double maxdiffA[p + 1];
int icol, jrow;

int main() {
    omp_set_num_threads(p);

    double h = 1.0 / (n + 1);

    double start = omp_get_wtime();

    #pragma omp parallel for private(icol) shared(x, y, v, _new)
    for (icol = 0; icol <= n + 1; ++icol) {
        x[icol] = y[icol] = icol * h;

        _new[icol][0] = v[icol][0] = 6 - 2 * x[icol];

        _new[n + 1][icol] = v[n + 1][icol] = 4 - 2 * y[icol];

        _new[icol][n + 1] = v[icol][n + 1] = 3 - x[icol];

        _new[0][icol] = v[0][icol] = 6 - 3 * y[icol];
    }

    const double eps = 0.01;


    #pragma omp parallel private(icol, jrow) shared(_new, v, maxdiffA)
    {
        while (true) { //for [iters=1 to maxiters by 2]
            #pragma omp single
            for (int i = 0; i < p; i++) maxdiffA[i] = 0;
            #pragma omp for
            for (icol = 1; icol <= n; icol++)
                for (jrow = 1; jrow <= n; jrow++)
                    _new[icol][jrow] =
                            (v[icol - 1][jrow] + v[icol + 1][jrow] + v[icol][jrow - 1] + v[icol][jrow + 1]) / 4;
            #pragma omp for
            for (icol = 1; icol <= n; icol++)
                for (jrow = 1; jrow <= n; jrow++)
                    v[icol][jrow] = (_new[icol - 1][jrow] + _new[icol + 1][jrow] + _new[icol][jrow - 1] +
                                     _new[icol][jrow + 1]) / 4;

            #pragma omp for
            for (icol = 1; icol <= n; icol++)
                for (jrow = 1; jrow <= n; jrow++)
                    maxdiffA[omp_get_thread_num()] = max(maxdiffA[omp_get_thread_num()],
                                                         fabs(_new[icol][jrow] - v[icol][jrow]));

            #pragma omp barrier

            double maxdiff = 0.0;
            for (int k = 0; k < p; ++k) {
                maxdiff = max(maxdiff, maxdiffA[k]);
            }


            if (maxdiff < eps)
                break;
            #pragma omp single
            std::cout << maxdiff << std::endl;
        }
    }
    double end = omp_get_wtime();
    printf("start = %.16lf\nend = %.16lf\ndiff = %.16lf\n", start, end, end - start);

    return 0;
}

输出

1.12454
<106 iterations here>
0.0100436
start = 1527366091.3069999217987061
end = 1527366110.8169999122619629
diff = 19.5099999904632568

但如果我删除

#pragma omp single
std::cout << maxdiff << std::endl;

该程序似乎运行无限长或我

start = 1527368219.8810000419616699
end = 1527368220.5710000991821289
diff = 0.6900000572204590

为什么会这样？

Answer 1

您在while循环的开头覆盖maxdiffA - 这必须与最后读取maxdiffA隔离以检查条件。否则，在另一个线程有机会读取它们之前，一个线程可能已经重置了这些值。由于omp single结构末尾的隐式屏障，循环结束时的omp single构造充当隔离。然而，在omp single构造的开头没有障碍。此外＆＃34;大量代码＆＃34; 不是安全屏障。因此，如果没有有效的隐式屏障，则必须使用#pragma omp barrier保护重置代码的输入。

这就是说我强烈建议重新构造代码以使共享退出条件也在single构造中计算。这使得更清楚的是所有线程进程同时退出while - 循环。否则代码定义不明确。

删除单个构造会导致执行错误

1 个答案: