1  void abort (void);
       2  
       3  void parloop (int N)
       4  {
       5    int i, j;
       6    int x[500][500];
       7  
       8    for (i = 0; i < N; i++)
       9      for (j = 0; j < N; j++)
      10        x[i][j] = i + j + 3;
      11  
      12    for (i = 0; i < N; i++)
      13      for (j = 0; j < N; j++)
      14        if (x[i][j] != i + j + 3)
      15  	abort ();
      16  }
      17  
      18  int main(void)
      19  {
      20    parloop(500);
      21  
      22    return 0;
      23  }
      24  
      25  /* Check that parallel code generation part make the right answer.  */
      26  /* { dg-final { scan-tree-dump-times "2 loops carried no dependency" 1 "graphite" } } */
      27  /* { dg-final { scan-tree-dump-times "loopfn" 4 "optimized" } } */