multithreading - 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，我希望 6 个线程专用于 task1，另外 2 个专用于 task2。

如何使用 OpenMP 实现这一点？

Answer 1

这是 OpenMP nested parallelism 的工作，从 OpenMP 3 开始:您可以使用 OpenMP tasks启动两个独立的任务，然后在这些任务中，有使用适当线程数的并行部分。

举个简单的例子:

#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 8 核(16 个硬件线程)节点上运行，

$ 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

已更新:我更改了上面的内容以包含线程祖先；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 个子线程中的线程 1”与“0 的 11 个子线程中的线程 1” ”)。

来自OpenMP standard ，S.3.2.4，“omp_get_thread_num 例程返回线程号，在当前团队中，调用线程。”，以及来自第 2.5 节的“当一个线程遇到并行构造，创建一组线程以执行并行区域 [...] 遇到并行构造的线程成为新团队的主线程，线程号为零新平行区域的持续时间。”

也就是说，在每个(嵌套的)并行区域中，创建线程组，线程 ID 从零开始；但仅仅因为这些 ID 在团队中重叠并不意味着它们是相同的线程。在这里，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 密集型工作，您还会通过监视工具看到确实有 16 个事件线程，而不仅仅是 11 个。

它们是团队本地线程编号而不是全局唯一线程编号的原因很简单；在可能发生嵌套和动态并行性的环境中，几乎不可能跟踪全局唯一的线程数。假设有三组线程，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有空隙？我们是否中断所有线程以更改它们的全局编号？如果启动了一个新团队，有 7 个线程怎么办？我们是从 6 开始并具有重叠的线程 ID，还是从 16 开始并在编号中留有间隙？