1  void abort (void);
       2  
       3  int x[10000000];
       4  
       5  void parloop (int N)
       6  {
       7    int i;
       8  
       9    for (i = 0; i < N; i++)
      10      x[i] = i + 1;
      11  
      12    for (i = 0; i < N; i++)
      13      x[i] = i + 3;
      14  
      15    for (i = 0; i < N; i++)
      16      {
      17        if (x[i] != i + 3)
      18  	abort ();
      19      }
      20  }
      21  
      22  int main(void)
      23  {
      24    parloop(10000000);
      25  
      26    return 0;
      27  }
      28  
      29  /* Check that parallel code generation part make the right answer.  */
      30  /* { dg-final { scan-tree-dump-times "2 loops carried no dependency" 1 "graphite" } } */
      31  /* { dg-final { scan-tree-dump-times "loopfn" 8 "optimized" } } */