1 void abort (void);
2 
3 int x[10000000];
4 
parloop(int N)5 void parloop (int N)
6 {
7   int i;
8 
9   for (i = 0; i < N; i++)
10     x[i] = i + 3;
11 
12   for (i = 0; i < N; i++)
13     {
14       if (x[i] != i + 3)
15 	abort ();
16     }
17 }
18 
main(void)19 int main(void)
20 {
21   parloop(10000000);
22 
23   return 0;
24 }
25 
26 /* Check that parallel code generation part make the right answer.  */
27 /* { dg-final { scan-tree-dump-times "1 loops carried no dependency" 1 "graphite" } } */
28 /* { dg-final { cleanup-tree-dump "graphite" } } */
29 /* { dg-final { scan-tree-dump-times "loopfn" 5 "optimized" } } */
30 /* { dg-final { cleanup-tree-dump "parloops" } } */
31 /* { dg-final { cleanup-tree-dump "optimized" } } */
32