OpenMP中的SECTIONS指令如何分配工作? [英] How does the SECTIONS directive in OpenMP distribute work?

问题描述

在使用omp sections的OpenMP中，线程将分配到节内的块，还是将每个线程分配给每个节?

In OpenMP when using omp sections, will the threads be distributed to the blocks inside the sections, or will each thread be assigned to each sections?

nthreads == 3时:

#pragma omp sections
{
    #pragma omp section
    { 
        printf ("id = %d, \n", omp_get_thread_num());
    }

    #pragma omp section
    { 
        printf ("id = %d, \n", omp_get_thread_num());
    }
}

输出:

id=1
id=1

但是当我执行以下代码时:

But when I execute the following code:

#pragma omp sections
{
    #pragma omp section
    { 
        printf ("id = %d, \n", omp_get_thread_num());
    }

    #pragma omp section
    { 
        printf ("id = %d, \n", omp_get_thread_num());
    }
}

#pragma omp sections
{
    #pragma omp section
    { 
        printf ("id = %d, \n", omp_get_thread_num());
    }

    #pragma omp section
    { 
        printf ("id = %d, \n", omp_get_thread_num());
    }
}

输出:

id=1
id=1

id=2
id=2

从这些输出中，我无法理解OpenMP中节的概念.

From these output I can't understand what the concept of sections is in OpenMP.

推荐答案

由OP发布的代码将永远不会并行执行，因为没有出现parallel关键字. OP的id不同于0的事实表明，他的代码可能嵌入在并行指令中.但是，从他的职位上并不清楚，这可能会使初学者感到困惑.

The code posted by the OP will never execute in parallel, because the parallel keyword does not appear. The fact that the OP got ids different from 0 shows that probably his code was embedded in a parallel directive. However, this is not clear from his post, and might confuse beginners.

最明智的例子是(对于由OP发布的第一个例子):

The minimum sensible example is (for the first example posted by the OP):

#pragma omp parallel sections
{
    #pragma omp section
    { 
        printf ("id = %d, \n", omp_get_thread_num());
    }

    #pragma omp section
    { 
        printf ("id = %d, \n", omp_get_thread_num());
    }
}

在我的机器上，这会打印

On my machine, this prints

id = 0,
id = 1,

表明这两个部分是由不同的线程执行的.

showing that the two sections are being executed by different threads.

值得注意的是，此代码不能比两个线程提取更多的并行度:如果执行的线程更多，则其他线程没有任何工作要做，只会闲置.

It's worth noting that however this code can not extract more parallelism than two threads: if it is executed with more threads, the other threads don't have any work to do and will just sit down idle.

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