c++ - "final"指令的 "task"子句是否正确？

Question

我正在尝试了解如何制作 final task 的条款实际上有效，我在 this 中找到了以下代码示例链接。

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

#define THRESHOLD 5

int fib(int n)
{
  int i, j;

  if (n<2)
  return n;

  #pragma omp task shared(i) firstprivate(n) final(n <= THRESHOLD)
  i=fib(n-1);

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

  #pragma omp taskwait
  return i+j;
}


int main()
{
  int n = 30;
  omp_set_dynamic(0);
  omp_set_num_threads(4);

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

但是当我检查 n 的值时通过调试，我注意到即使在条件 n <= THRESHOLD 时，递归函数显然仍在被调用。是真的。所以，我的问题是，是我误解了什么还是代码有问题？

Answer 1

当您调用 final 时发生的事情是，从这个任务中派生出来的任务不会从中创建一个新任务。

来自 Specification :

[§ 2.10.1 task Construct] When a final clause is present on a task construct and the final clause expression evaluates to true, the generated task will be a final task. All task constructs encountered during execution of a final task will generate final and included tasks. Note that the use of a variable in a final clause expression of a task construct causes an implicit reference to the variable in all enclosing constructs.

最后的任务是:

A task that forces all of its child tasks to become final and included tasks.

包含的任务是:

A task for which execution is sequentially included in the generating task region. That is, an included task is undeferred and executed immediately by the encountering thread.

在您的情况下，这意味着:如果在此任务中满足 THREASHOLD，则会生成 n-1 和 n-2 生成的新任务，并且不会生成这些任务递归调用时的新任务，但仍然以“老式”函数调用方式调用函数。