Question

OpenMP之间有什么区别：

#pragma omp parallel sections
    {
        #pragma omp section
        {
           fct1();
        }
        #pragma omp section
        {
           fct2();
        }
}

和：

#pragma omp parallel 
{
    #pragma omp single
    {
       #pragma omp task
       fct1();
       #pragma omp task
       fct2();
    }
}

我不确定第二个代码是否正确......

Answer 1

任务和部分之间的区别在于代码执行的时间范围。节被包含在sections构造中，并且（除非指定了nowait子句）线程将不会离开它，直到所有节都被执行：

                 [    sections     ]
Thread 0: -------< section 1 >---->*------
Thread 1: -------< section 2      >*------
Thread 2: ------------------------>*------
...                                *
Thread N-1: ---------------------->*------

这里N个线程遇到一个sections构造有两个部分，第二个比第一个占用更多的时间。前两个线程各执行一个部分。其他N-2线程只是在sections结构末尾的隐式屏障处等待（此处显示为*）。

任务在所谓的任务调度点处尽可能排队并执行。在某些情况下，可以允许运行时在线程之间移动任务，即使在它们的生命周期中也是如此。这些任务被称为解开，一个解开的任务可能会在一个线程中开始执行，然后在某个调度点，它可能会被运行时迁移到另一个线程。

但是，任务和部分在很多方面都相似。例如，以下两个代码片段实现了基本相同的结果：

// sections
...
#pragma omp sections
{
   #pragma omp section
   foo();
   #pragma omp section
   bar();
}
...

// tasks
...
#pragma omp single nowait
{
   #pragma omp task
   foo();
   #pragma omp task
   bar();
}
#pragma omp taskwait
...

taskwait与barrier非常相似，但对于任务 - 它确保当前执行流程暂停，直到所有排队任务都已执行。它是一个调度点，即它允许线程处理任务。需要single构造，以便仅由一个线程创建任务。如果没有single构造，则每个任务将被创建num_threads次，这可能不是人们想要的。 nowait构造中的single子句指示其他线程不等待single构造执行（即删除single构造末尾的隐式障碍）。所以他们立即点击taskwait并开始处理任务。

taskwait是为清晰起见而在此处显示的显式调度点。还有隐式调度点，最明显的是在屏障同步内部，无论是显式还是隐式。因此，上面的代码也可以简单地写成：

// tasks
...
#pragma omp single
{
   #pragma omp task
   foo();
   #pragma omp task
   bar();
}
...

如果有三个线程，可能会发生以下情况：

               +--+-->[ task queue ]--+
               |  |                   |
               |  |       +-----------+
               |  |       |
Thread 0: --< single >-|  v  |-----
Thread 1: -------->|< foo() >|-----
Thread 2: -------->|< bar() >|-----

在此| ... |中显示的是调度点的操作（taskwait指令或隐式屏障）。基本上，线程1和2暂停它们正在执行的操作，并从队列开始处理任务。处理完所有任务后，线程将恢复正常的执行流程。请注意，线程1和2可能会在线程0退出single构造之前到达调度点，因此左|无需对齐（这在上图中表示。）

也可能发生线程1能够完成处理foo()任务并在其他线程能够请求任务之前请求另一个任务。因此，foo()和bar()都可能由同一个线程执行：

               +--+-->[ task queue ]--+
               |  |                   |
               |  |      +------------+
               |  |      |
Thread 0: --< single >-| v             |---
Thread 1: --------->|< foo() >< bar() >|---
Thread 2: --------------------->|      |---

如果线程2来得太晚，单挑线程也可能执行第二个任务：

               +--+-->[ task queue ]--+
               |  |                   |
               |  |      +------------+
               |  |      |
Thread 0: --< single >-| v < bar() >|---
Thread 1: --------->|< foo() >      |---
Thread 2: ----------------->|       |---

在某些情况下，编译器或OpenMP运行时甚至可能完全绕过任务队列并以串行方式执行任务：

Thread 0: --< single: foo(); bar() >*---
Thread 1: ------------------------->*---
Thread 2: ------------------------->*---

如果区域代码中没有任何任务调度点，则OpenMP运行时可能会在其认为合适时启动任务。例如，可能会延迟所有任务，直到达到parallel区域末尾的障碍。

Answer 2

我不是OpenMP专家，但是尝试同时使用task和sections在我的机器上测试fib序列

部分

int fib(int n)
{
    int i, j;
    if (n < 2)
        return n;
    else
    {
#pragma omp parallel sections       
{
#pragma omp section             
{
                i = fib(n - 1);
            }
#pragma omp section             
{
                j = fib(n - 2);
            }
        }
        printf("Current int %d is on thread %d \n", i + j, omp_get_thread_num());
        return i + j;
    }
}

int main()
{
    int n = 10;

#pragma omp parallel shared(n)  {
#pragma omp single      {
            printf("%d\n", omp_get_num_threads());
            printf("fib(%d) = %d\n", n, fib(n));
        }
    }
}

任务

#include <stdio.h>
#include <omp.h>
int fib(int n)
{
  int i, j;
  if (n<2)
    return n;
  else
    {
       #pragma omp task shared(i) firstprivate(n)
       i=fib(n-1);

       #pragma omp task shared(j) firstprivate(n)
       j=fib(n-2);

       #pragma omp taskwait
    printf("Current int %d is on thread %d \n", i + j, omp_get_thread_num());
       return i+j;
    }
}

int main()
{
  int n = 10;

  #pragma omp parallel shared(n)
  {
    #pragma omp single
    {
    printf("%d\n", omp_get_num_threads());
        printf ("fib(%d) = %d\n", n, fib(n));
    }
  }
}

部分结果：

12
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 1 is on thread 0
Current int 3 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 5 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 1 is on thread 0
Current int 3 is on thread 0
Current int 8 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 1 is on thread 0
Current int 3 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 5 is on thread 0
Current int 13 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 1 is on thread 0
Current int 3 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 5 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 1 is on thread 0
Current int 3 is on thread 0
Current int 8 is on thread 0
Current int 21 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 1 is on thread 0
Current int 3 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 5 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 1 is on thread 0
Current int 3 is on thread 0
Current int 8 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 1 is on thread 0
Current int 3 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 5 is on thread 0
Current int 13 is on thread 0
Current int 34 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 1 is on thread 0
Current int 3 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 5 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 1 is on thread 0
Current int 3 is on thread 0
Current int 8 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 1 is on thread 0
Current int 3 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 5 is on thread 0
Current int 13 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 1 is on thread 0
Current int 3 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 5 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 1 is on thread 0
Current int 3 is on thread 0
Current int 8 is on thread 0
Current int 21 is on thread 0
Current int 55 is on thread 4
fib(10) = 55

任务结果：

12
Current int 1 is on thread 3
Current int 2 is on thread 3
Current int 1 is on thread 8
Current int 2 is on thread 8
Current int 1 is on thread 8
Current int 1 is on thread 4
Current int 1 is on thread 11
Current int 1 is on thread 11
Current int 2 is on thread 11
Current int 3 is on thread 11
Current int 1 is on thread 11
Current int 2 is on thread 11
Current int 1 is on thread 11
Current int 1 is on thread 11
Current int 2 is on thread 11
Current int 3 is on thread 11
Current int 1 is on thread 11
Current int 2 is on thread 11
Current int 1 is on thread 11
Current int 1 is on thread 11
Current int 2 is on thread 11
Current int 3 is on thread 11
Current int 5 is on thread 11
Current int 8 is on thread 11
Current int 1 is on thread 8
Current int 2 is on thread 8
Current int 3 is on thread 8
Current int 5 is on thread 8
Current int 13 is on thread 8
Current int 1 is on thread 7
Current int 2 is on thread 7
Current int 1 is on thread 7
Current int 1 is on thread 7
Current int 1 is on thread 0
Current int 1 is on thread 0
Current int 2 is on thread 0
Current int 3 is on thread 0
Current int 1 is on thread 1
Current int 1 is on thread 6
Current int 2 is on thread 6
Current int 1 is on thread 9
Current int 2 is on thread 9
Current int 1 is on thread 2
Current int 2 is on thread 7
Current int 3 is on thread 7
Current int 5 is on thread 7
Current int 2 is on thread 5
Current int 5 is on thread 5
Current int 1 is on thread 5
Current int 2 is on thread 5
Current int 1 is on thread 5
Current int 1 is on thread 5
Current int 2 is on thread 5
Current int 3 is on thread 5
Current int 1 is on thread 5
Current int 2 is on thread 5
Current int 1 is on thread 5
Current int 1 is on thread 5
Current int 2 is on thread 5
Current int 3 is on thread 5
Current int 5 is on thread 5
Current int 1 is on thread 5
Current int 2 is on thread 5
Current int 1 is on thread 11
Current int 2 is on thread 11
Current int 1 is on thread 8
Current int 2 is on thread 8
Current int 5 is on thread 8
Current int 3 is on thread 1
Current int 8 is on thread 1
Current int 21 is on thread 1
Current int 1 is on thread 10
Current int 3 is on thread 10
Current int 8 is on thread 0
Current int 1 is on thread 4
Current int 3 is on thread 4
Current int 1 is on thread 9
Current int 3 is on thread 9
Current int 8 is on thread 9
Current int 3 is on thread 2
Current int 5 is on thread 3
Current int 13 is on thread 3
Current int 5 is on thread 6
Current int 13 is on thread 7
Current int 8 is on thread 10
Current int 21 is on thread 10
Current int 34 is on thread 3
Current int 55 is on thread 1
fib(10) = 55

在分配计算资源时，任务似乎比部分要聪明得多

--------------------------------- EDIT ----------------- ------------

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部分和任务openmp之间的区别

2 个答案: