Question

#pragma omp parallel shared(newgrid,grid,maxdiff,epsilon) private(i,j,diff,maxdiff1,grid0,num_iter)
{   
    do{
    #pragma omp critical
    maxdiff = 0;
    //num_iter=num_iter+1;
    #pragma omp for
    for (i=1;i<n;i++)
    {   
        maxdiff1 = 0;
        for (j=1;j<n;j++){
            grid0 = grid[i+(n+1)*j];
            newgrid[i+(n+1)*j] = (grid[i-1+(n+1)*j]+grid[i+1+(n+1)*j]+grid[i+(n+1)*(j-1)]+grid[i+(n+1)*(j+1)]-h*h*f[i-1+(n-1)*(j-1)])/4;
            diff = fabs(newgrid[i+(n+1)*j]-grid0);
            grid[i+(n+1)*j] = newgrid[i+(n+1)*j];
            if (diff>maxdiff1) 
                maxdiff1 = diff;
        }
        #pragma omp critical
        if (maxdiff1 > maxdiff) maxdiff = maxdiff1; 
    }
    }while(maxdiff>=epsilon);
}

初始化openmp每个do {} while（）循环的内容我在循环之前创建theads并在线程之间并行化i循环。问题是它在2个线程上无法正常工作，但正确地在1上，并且它不时卡住。

Answer 1

当您写信至newgrid时，您有竞争条件。例如，当一个线程的i与另一个线程的i-1相同时（同样适用于j）。如果要获得正确的结果，请在写入newgrid时使用临界区。

    do {
        maxdiff = 0;
        #pragma omp for private(i,j, diff, maxdiff1, grid0)
        for (i=1;i<n;i++) {   
            maxdiff1 = 0;
            for (j=1;j<n;j++) {
                grid0 = grid[i+(n+1)*j];
                #pragma omp critical
                newgrid[i+(n+1)*j] = (grid[i-1+(n+1)*j]+grid[i+1+(n+1)*j]+grid[i+(n+1)*(j-1)]+grid[i+(n+1)*(j+1)]-h*h*f[i-1+(n-1)*(j-1)])/4;
                diff = fabs(newgrid[i+(n+1)*j]-grid0);
                grid[i+(n+1)*j] = newgrid[i+(n+1)*j];
                if (diff>maxdiff1) 
                    maxdiff1 = diff;
            }
            #pragma omp critical
            if (maxdiff1 > maxdiff) maxdiff = maxdiff1; 
        }
    }while(maxdiff>=epsilon);

openmp线程和循环

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