2 use crate::mem::{self, MaybeUninit};
5 /// Rotates the range `[mid-left, mid+right)` such that the element at `mid` becomes the first
6 /// element. Equivalently, rotates the range `left` elements to the left or `right` elements to the
11 /// The specified range must be valid for reading and writing.
15 /// Algorithm 1 is used for small values of `left + right` or for large `T`. The elements are moved
16 /// into their final positions one at a time starting at `mid - left` and advancing by `right` steps
17 /// modulo `left + right`, such that only one temporary is needed. Eventually, we arrive back at
18 /// `mid - left`. However, if `gcd(left + right, right)` is not 1, the above steps skipped over
19 /// elements. For example:
21 /// left = 10, right = 6
22 /// the `^` indicates an element in its final place
23 /// 6 7 8 9 10 11 12 13 14 15 . 0 1 2 3 4 5
24 /// after using one step of the above algorithm (The X will be overwritten at the end of the round,
25 /// and 12 is stored in a temporary):
26 /// X 7 8 9 10 11 6 13 14 15 . 0 1 2 3 4 5
28 /// after using another step (now 2 is in the temporary):
29 /// X 7 8 9 10 11 6 13 14 15 . 0 1 12 3 4 5
31 /// after the third step (the steps wrap around, and 8 is in the temporary):
32 /// X 7 2 9 10 11 6 13 14 15 . 0 1 12 3 4 5
34 /// after 7 more steps, the round ends with the temporary 0 getting put in the X:
35 /// 0 7 2 9 4 11 6 13 8 15 . 10 1 12 3 14 5
38 /// Fortunately, the number of skipped over elements between finalized elements is always equal, so
39 /// we can just offset our starting position and do more rounds (the total number of rounds is the
40 /// `gcd(left + right, right)` value). The end result is that all elements are finalized once and
43 /// Algorithm 2 is used if `left + right` is large but `min(left, right)` is small enough to
44 /// fit onto a stack buffer. The `min(left, right)` elements are copied onto the buffer, `memmove`
45 /// is applied to the others, and the ones on the buffer are moved back into the hole on the
46 /// opposite side of where they originated.
48 /// Algorithms that can be vectorized outperform the above once `left + right` becomes large enough.
49 /// Algorithm 1 can be vectorized by chunking and performing many rounds at once, but there are too
50 /// few rounds on average until `left + right` is enormous, and the worst case of a single
51 /// round is always there. Instead, algorithm 3 utilizes repeated swapping of
52 /// `min(left, right)` elements until a smaller rotate problem is left.
55 /// left = 11, right = 4
56 /// [4 5 6 7 8 9 10 11 12 13 14 . 0 1 2 3]
57 /// ^ ^ ^ ^ ^ ^ ^ ^ swapping the right most elements with elements to the left
58 /// [4 5 6 7 8 9 10 . 0 1 2 3] 11 12 13 14
59 /// ^ ^ ^ ^ ^ ^ ^ ^ swapping these
60 /// [4 5 6 . 0 1 2 3] 7 8 9 10 11 12 13 14
61 /// we cannot swap any more, but a smaller rotation problem is left to solve
63 /// when `left < right` the swapping happens from the left instead.
64 pub unsafe fn ptr_rotate<T>(mut left: usize, mut mid: *mut T, mut right: usize) {
65 type BufType = [usize; 32];
66 if mem::size_of::<T>() == 0 {
70 // N.B. the below algorithms can fail if these cases are not checked
71 if (right == 0) || (left == 0) {
74 if (left + right < 24) || (mem::size_of::<T>() > mem::size_of::<[usize; 4]>()) {
76 // Microbenchmarks indicate that the average performance for random shifts is better all
77 // the way until about `left + right == 32`, but the worst case performance breaks even
78 // around 16. 24 was chosen as middle ground. If the size of `T` is larger than 4
79 // `usize`s, this algorithm also outperforms other algorithms.
80 let x = mid.sub(left);
81 // beginning of first round
82 let mut tmp: T = x.read();
84 // `gcd` can be found before hand by calculating `gcd(left + right, right)`,
85 // but it is faster to do one loop which calculates the gcd as a side effect, then
86 // doing the rest of the chunk
88 // benchmarks reveal that it is faster to swap temporaries all the way through instead
89 // of reading one temporary once, copying backwards, and then writing that temporary at
90 // the very end. This is possibly due to the fact that swapping or replacing temporaries
91 // uses only one memory address in the loop instead of needing to manage two.
93 tmp = x.add(i).replace(tmp);
94 // instead of incrementing `i` and then checking if it is outside the bounds, we
95 // check if `i` will go outside the bounds on the next increment. This prevents
96 // any wrapping of pointers or `usize`.
100 // end of first round
104 // this conditional must be here if `left + right >= 15`
112 // finish the chunk with more rounds
113 for start in 1..gcd {
114 tmp = x.add(start).read();
117 tmp = x.add(i).replace(tmp);
121 x.add(start).write(tmp);
130 // `T` is not a zero-sized type, so it's okay to divide by its size.
131 } else if cmp::min(left, right) <= mem::size_of::<BufType>() / mem::size_of::<T>() {
133 // The `[T; 0]` here is to ensure this is appropriately aligned for T
134 let mut rawarray = MaybeUninit::<(BufType, [T; 0])>::uninit();
135 let buf = rawarray.as_mut_ptr() as *mut T;
136 let dim = mid.sub(left).add(right);
138 ptr::copy_nonoverlapping(mid.sub(left), buf, left);
139 ptr::copy(mid, mid.sub(left), right);
140 ptr::copy_nonoverlapping(buf, dim, left);
142 ptr::copy_nonoverlapping(mid, buf, right);
143 ptr::copy(mid.sub(left), dim, left);
144 ptr::copy_nonoverlapping(buf, mid.sub(left), right);
147 } else if left >= right {
149 // There is an alternate way of swapping that involves finding where the last swap
150 // of this algorithm would be, and swapping using that last chunk instead of swapping
151 // adjacent chunks like this algorithm is doing, but this way is still faster.
153 ptr::swap_nonoverlapping(mid.sub(right), mid, right);
154 mid = mid.sub(right);
161 // Algorithm 3, `left < right`
163 ptr::swap_nonoverlapping(mid.sub(left), mid, left);