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void abort (void);
void parloop (int N)
{
int i, j;
int x[500][500];
for (i = 0; i < N; i++)
for (j = 0; j < N; j++)
x[i][j] = i + j + 3;
for (i = 0; i < N; i++)
for (j = 0; j < N; j++)
if (x[i][j] != i + j + 3)
abort ();
}
int main(void)
{
parloop(500);
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 { scan-tree-dump-times "loopfn" 4 "optimized" } } */
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