2 Checks for use of `.skip(x).next()` on iterators.
9 let some_vec = vec![0, 1, 2, 3];
10 let bad_vec = some_vec.iter().skip(3).next();
11 let bad_slice = &some_vec[..].iter().skip(3).next();
13 The correct use would be:
15 let some_vec = vec![0, 1, 2, 3];
16 let bad_vec = some_vec.iter().nth(3);
17 let bad_slice = &some_vec[..].iter().nth(3);