OpenMP:将所有线程分成不同的组

Question

我想将所有线程分成两个不同的组，因为我有两个并行任务要异步运行。例如，如果总共有8个线程可用，我将喜欢6个专用于task1的线程，另2个专用于task2的线程。

如何使用OpenMP实现这一目标？

Answer 1

这是OpenMP嵌套并行的一项任务，从OpenMP 3开始:您可以使用OpenMP任务启动两个独立的任务，然后在这些任务中有使用适当数量线程的并行部分。

作为一个简单的例子：

#include <stdio.h>
#include <omp.h>

int main(int argc, char **argv) {

    omp_set_nested(1);   /* make sure nested parallism is on */
    int nprocs = omp_get_num_procs();
    int nthreads1 = nprocs/3;
    int nthreads2 = nprocs - nthreads1;

    #pragma omp parallel default(none) shared(nthreads1, nthreads2) num_threads(2)
    #pragma omp single
    {
        #pragma omp task
        #pragma omp parallel for num_threads(nthreads1)
        for (int i=0; i<16; i++)
            printf("Task 1: thread %d of the %d children of %d: handling iter %d\n",
                        omp_get_thread_num(), omp_get_team_size(2),
                        omp_get_ancestor_thread_num(1), i);
        #pragma omp task
        #pragma omp parallel for num_threads(nthreads2)
        for (int j=0; j<16; j++)
            printf("Task 2: thread %d of the %d children of %d: handling iter %d\n",
                        omp_get_thread_num(), omp_get_team_size(2),
                        omp_get_ancestor_thread_num(1), j);
    }

    return 0;
}

在一个8核(16个硬件线程)节点上运行它，

$ gcc -fopenmp nested.c -o nested -std=c99
$ ./nested
Task 2: thread 3 of the 11 children of 0: handling iter 6
Task 2: thread 3 of the 11 children of 0: handling iter 7
Task 2: thread 1 of the 11 children of 0: handling iter 2
Task 2: thread 1 of the 11 children of 0: handling iter 3
Task 1: thread 2 of the 5 children of 1: handling iter 8
Task 1: thread 2 of the 5 children of 1: handling iter 9
Task 1: thread 2 of the 5 children of 1: handling iter 10
Task 1: thread 2 of the 5 children of 1: handling iter 11
Task 2: thread 6 of the 11 children of 0: handling iter 12
Task 2: thread 6 of the 11 children of 0: handling iter 13
Task 1: thread 0 of the 5 children of 1: handling iter 0
Task 1: thread 0 of the 5 children of 1: handling iter 1
Task 1: thread 0 of the 5 children of 1: handling iter 2
Task 1: thread 0 of the 5 children of 1: handling iter 3
Task 2: thread 5 of the 11 children of 0: handling iter 10
Task 2: thread 5 of the 11 children of 0: handling iter 11
Task 2: thread 0 of the 11 children of 0: handling iter 0
Task 2: thread 0 of the 11 children of 0: handling iter 1
Task 2: thread 2 of the 11 children of 0: handling iter 4
Task 2: thread 2 of the 11 children of 0: handling iter 5
Task 1: thread 1 of the 5 children of 1: handling iter 4
Task 2: thread 4 of the 11 children of 0: handling iter 8
Task 2: thread 4 of the 11 children of 0: handling iter 9
Task 1: thread 3 of the 5 children of 1: handling iter 12
Task 1: thread 3 of the 5 children of 1: handling iter 13
Task 1: thread 3 of the 5 children of 1: handling iter 14
Task 2: thread 7 of the 11 children of 0: handling iter 14
Task 2: thread 7 of the 11 children of 0: handling iter 15
Task 1: thread 1 of the 5 children of 1: handling iter 5
Task 1: thread 1 of the 5 children of 1: handling iter 6
Task 1: thread 1 of the 5 children of 1: handling iter 7
Task 1: thread 3 of the 5 children of 1: handling iter 15

更新的：我已经将上面的内容更改为包含线程祖先；出现了混淆，因为(例如)打印了两个“线程1”--这里我还打印了祖先(例如，“1的5个子线程中的一个线程1与0的11个子线程1中的线程1”)。

在OpenMP标准 S.3.2.4中，“omp_get_thread_num例程返回调用线程的当前团队中的线程号”，以及第2.5节，“当线程遇到并行构造时，创建一个线程组来执行并行区域.遇到并行构造的线程将成为新团队的主线程，在新的并行区域的持续时间内线程数为零。“

也就是说，在这些(嵌套的)并行区域中，创建的线程组的线程is从零开始；但仅仅因为这些is在团队中重叠并不意味着它们是相同的线程。这里我强调了，通过打印它们的祖先编号，但是如果线程正在执行CPU密集型的工作，您也会看到使用监视工具确实有16个活动线程，而不仅仅是11个。

它们是团队本地线程号而不是全局唯一线程号的原因非常简单；在可以发生嵌套和动态并行的环境中，几乎不可能跟踪全局唯一的线程号。假设有三个线程组，编号为0..5、6、..10和11.15，中间组完成。我们是否在线程编号中留下空白？我们是否中断所有线程以更改它们的全局编号？如果一个有7个线程的新团队启动了呢？我们是从6开始，有重叠的线程in，还是从16开始，在编号中留下空白？