#define N 1500
int x[N][N], y[N];
int foo(void)
{
int i, j;
for (i = 0; i < N; i++)
y[i] = i;
for (j = 0; j < N; j++)
if (j > 500)
x[i][j] = i + j + 3;
y[j] = i*j + 10;
}
else
x[i][j] = x[i][j]*3;
return x[2][5]*y[8];
int main(void)
foo();
return 0;
/* Check that parallel code generation part make the right answer. */
/* { dg-final { scan-tree-dump-times "2 loops carried no dependency" 1 "graphite" } } */
/* { dg-final { cleanup-tree-dump "graphite" } } */
/* { dg-final { scan-tree-dump-times "loopfn.0" 5 "optimized" } } */
/* { dg-final { scan-tree-dump-times "loopfn.1" 5 "optimized" } } */
/* { dg-final { cleanup-tree-dump "parloops" } } */
/* { dg-final { cleanup-tree-dump "optimized" } } */