这是一份工作OpenMP 嵌套并行性 http://docs.oracle.com/cd/E19205-01/819-5270/aewbc/index.html，从 OpenMP 3 开始：您可以使用OpenMP 任务 http://docs.oracle.com/cd/E19205-01/820-7883/6nj43o69j/index.html启动两个独立的任务，然后在这些任务中使用适当数量的线程的并行部分。
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
Updated：我已经更改了上面的内容以包含线程祖先；出现了混乱，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 个子级的线程 1”与“0 的 11 个子级的线程 1”） ”）。
来自OpenMP 标准 http://www.openmp.org/mp-documents/OpenMP4.0.0.pdf，S.3.2.4，“omp_get_thread_num例程返回线程号，目前团队内，调用线程的。”，以及第 2.5 节，“当一个线程遇到并行构造时，会创建一组线程来
执行并行区域 [...] 遇到并行构造的线程
成为新团队的主线程，线程数为零
新并行区域的持续时间。”
也就是说，在每个（嵌套）并行区域内，都会创建线程组，其线程 ID 从零开始；但仅仅因为这些 ID 在团队内重叠并不意味着它们是相同的线程。在这里，我也通过打印它们的祖先编号来强调这一点，但如果线程正在执行 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；在可能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下间隙？我们是否要中断所有线程来更改它们的全局编号？如果启动一个新团队，有 7 个线程怎么办？我们是从 6 开始并有重叠的线程 ID，还是从 16 开始并在编号中留下间隙？
