1 files changed, 330 insertions, 0 deletions
diff --git a/drivers/md/dm-vdo/indexer/radix-sort.c b/drivers/md/dm-vdo/indexer/radix-sort.c
new file mode 100644
index 000000000000..66b8c706a1ef
--- /dev/null
+++ b/drivers/md/dm-vdo/indexer/radix-sort.c
@@ -0,0 +1,330 @@
+// SPDX-License-Identifier: GPL-2.0-only
+/*
+ * Copyright 2023 Red Hat
+ */
+
+#include "radix-sort.h"
+
+#include <linux/limits.h>
+#include <linux/types.h>
+
+#include "memory-alloc.h"
+#include "string-utils.h"
+
+/*
+ * This implementation allocates one large object to do the sorting, which can be reused as many
+ * times as desired. The amount of memory required is logarithmically proportional to the number of
+ * keys to be sorted.
+ */
+
+/* Piles smaller than this are handled with a simple insertion sort. */
+#define INSERTION_SORT_THRESHOLD 12
+
+/* Sort keys are pointers to immutable fixed-length arrays of bytes. */
+typedef const u8 *sort_key_t;
+
+/*
+ * The keys are separated into piles based on the byte in each keys at the current offset, so the
+ * number of keys with each byte must be counted.
+ */
+struct histogram {
+	/* The number of non-empty bins */
+	u16 used;
+	/* The index (key byte) of the first non-empty bin */
+	u16 first;
+	/* The index (key byte) of the last non-empty bin */
+	u16 last;
+	/* The number of occurrences of each specific byte */
+	u32 size[256];
+};
+
+/*
+ * Sub-tasks are manually managed on a stack, both for performance and to put a logarithmic bound
+ * on the stack space needed.
+ */
+struct task {
+	/* Pointer to the first key to sort. */
+	sort_key_t *first_key;
+	/* Pointer to the last key to sort. */
+	sort_key_t *last_key;
+	/* The offset into the key at which to continue sorting. */
+	u16 offset;
+	/* The number of bytes remaining in the sort keys. */
+	u16 length;
+};
+
+struct radix_sorter {
+	unsigned int count;
+	struct histogram bins;
+	sort_key_t *pile[256];
+	struct task *end_of_stack;
+	struct task insertion_list[256];
+	struct task stack[];
+};
+
+/* Compare a segment of two fixed-length keys starting at an offset. */
+static inline int compare(sort_key_t key1, sort_key_t key2, u16 offset, u16 length)
+{
+	return memcmp(&key1[offset], &key2[offset], length);
+}
+
+/* Insert the next unsorted key into an array of sorted keys. */
+static inline void insert_key(const struct task task, sort_key_t *next)
+{
+	/* Pull the unsorted key out, freeing up the array slot. */
+	sort_key_t unsorted = *next;
+
+	/* Compare the key to the preceding sorted entries, shifting down ones that are larger. */
+	while ((--next >= task.first_key) &&
+	       (compare(unsorted, next[0], task.offset, task.length) < 0))
+		next[1] = next[0];
+
+	/* Insert the key into the last slot that was cleared, sorting it. */
+	next[1] = unsorted;
+}
+
+/*
+ * Sort a range of key segments using an insertion sort. This simple sort is faster than the
+ * 256-way radix sort when the number of keys to sort is small.
+ */
+static inline void insertion_sort(const struct task task)
+{
+	sort_key_t *next;
+
+	for (next = task.first_key + 1; next <= task.last_key; next++)
+		insert_key(task, next);
+}
+
+/* Push a sorting task onto a task stack. */
+static inline void push_task(struct task **stack_pointer, sort_key_t *first_key,
+			     u32 count, u16 offset, u16 length)
+{
+	struct task *task = (*stack_pointer)++;
+
+	task->first_key = first_key;
+	task->last_key = &first_key[count - 1];
+	task->offset = offset;
+	task->length = length;
+}
+
+static inline void swap_keys(sort_key_t *a, sort_key_t *b)
+{
+	sort_key_t c = *a;
+	*a = *b;
+	*b = c;
+}
+
+/*
+ * Count the number of times each byte value appears in the arrays of keys to sort at the current
+ * offset, keeping track of the number of non-empty bins, and the index of the first and last
+ * non-empty bin.
+ */
+static inline void measure_bins(const struct task task, struct histogram *bins)
+{
+	sort_key_t *key_ptr;
+
+	/*
+	 * Subtle invariant: bins->used and bins->size[] are zero because the sorting code clears
+	 * it all out as it goes. Even though this structure is re-used, we don't need to pay to
+	 * zero it before starting a new tally.
+	 */
+	bins->first = U8_MAX;
+	bins->last = 0;
+
+	for (key_ptr = task.first_key; key_ptr <= task.last_key; key_ptr++) {
+		/* Increment the count for the byte in the key at the current offset. */
+		u8 bin = (*key_ptr)[task.offset];
+		u32 size = ++bins->size[bin];
+
+		/* Track non-empty bins. */
+		if (size == 1) {
+			bins->used += 1;
+			if (bin < bins->first)
+				bins->first = bin;
+
+			if (bin > bins->last)
+				bins->last = bin;
+		}
+	}
+}
+
+/*
+ * Convert the bin sizes to pointers to where each pile goes.
+ *
+ *   pile[0] = first_key + bin->size[0],
+ *   pile[1] = pile[0]  + bin->size[1], etc.
+ *
+ * After the keys are moved to the appropriate pile, we'll need to sort each of the piles by the
+ * next radix position. A new task is put on the stack for each pile containing lots of keys, or a
+ * new task is put on the list for each pile containing few keys.
+ *
+ * @stack: pointer the top of the stack
+ * @end_of_stack: the end of the stack
+ * @list: pointer the head of the list
+ * @pile: array for pointers to the end of each pile
+ * @bins: the histogram of the sizes of each pile
+ * @first_key: the first key of the stack
+ * @offset: the next radix position to sort by
+ * @length: the number of bytes remaining in the sort keys
+ *
+ * Return: UDS_SUCCESS or an error code
+ */
+static inline int push_bins(struct task **stack, struct task *end_of_stack,
+			    struct task **list, sort_key_t *pile[],
+			    struct histogram *bins, sort_key_t *first_key,
+			    u16 offset, u16 length)
+{
+	sort_key_t *pile_start = first_key;
+	int bin;
+
+	for (bin = bins->first; ; bin++) {
+		u32 size = bins->size[bin];
+
+		/* Skip empty piles. */
+		if (size == 0)
+			continue;
+
+		/* There's no need to sort empty keys. */
+		if (length > 0) {
+			if (size > INSERTION_SORT_THRESHOLD) {
+				if (*stack >= end_of_stack)
+					return UDS_BAD_STATE;
+
+				push_task(stack, pile_start, size, offset, length);
+			} else if (size > 1) {
+				push_task(list, pile_start, size, offset, length);
+			}
+		}
+
+		pile_start += size;
+		pile[bin] = pile_start;
+		if (--bins->used == 0)
+			break;
+	}
+
+	return UDS_SUCCESS;
+}
+
+int uds_make_radix_sorter(unsigned int count, struct radix_sorter **sorter)
+{
+	int result;
+	unsigned int stack_size = count / INSERTION_SORT_THRESHOLD;
+	struct radix_sorter *radix_sorter;
+
+	result = vdo_allocate_extended(struct radix_sorter, stack_size, struct task,
+				       __func__, &radix_sorter);
+	if (result != VDO_SUCCESS)
+		return result;
+
+	radix_sorter->count = count;
+	radix_sorter->end_of_stack = radix_sorter->stack + stack_size;
+	*sorter = radix_sorter;
+	return UDS_SUCCESS;
+}
+
+void uds_free_radix_sorter(struct radix_sorter *sorter)
+{
+	vdo_free(sorter);
+}
+
+/*
+ * Sort pointers to fixed-length keys (arrays of bytes) using a radix sort. The sort implementation
+ * is unstable, so the relative ordering of equal keys is not preserved.
+ */
+int uds_radix_sort(struct radix_sorter *sorter, const unsigned char *keys[],
+		   unsigned int count, unsigned short length)
+{
+	struct task start;
+	struct histogram *bins = &sorter->bins;
+	sort_key_t **pile = sorter->pile;
+	struct task *task_stack = sorter->stack;
+
+	/* All zero-length keys are identical and therefore already sorted. */
+	if ((count == 0) || (length == 0))
+		return UDS_SUCCESS;
+
+	/* The initial task is to sort the entire length of all the keys. */
+	start = (struct task) {
+		.first_key = keys,
+		.last_key = &keys[count - 1],
+		.offset = 0,
+		.length = length,
+	};
+
+	if (count <= INSERTION_SORT_THRESHOLD) {
+		insertion_sort(start);
+		return UDS_SUCCESS;
+	}
+
+	if (count > sorter->count)
+		return UDS_INVALID_ARGUMENT;
+
+	/*
+	 * Repeatedly consume a sorting task from the stack and process it, pushing new sub-tasks
+	 * onto the stack for each radix-sorted pile. When all tasks and sub-tasks have been
+	 * processed, the stack will be empty and all the keys in the starting task will be fully
+	 * sorted.
+	 */
+	for (*task_stack = start; task_stack >= sorter->stack; task_stack--) {
+		const struct task task = *task_stack;
+		struct task *insertion_task_list;
+		int result;
+		sort_key_t *fence;
+		sort_key_t *end;
+
+		measure_bins(task, bins);
+
+		/*
+		 * Now that we know how large each bin is, generate pointers for each of the piles
+		 * and push a new task to sort each pile by the next radix byte.
+		 */
+		insertion_task_list = sorter->insertion_list;
+		result = push_bins(&task_stack, sorter->end_of_stack,
+				   &insertion_task_list, pile, bins, task.first_key,
+				   task.offset + 1, task.length - 1);
+		if (result != UDS_SUCCESS) {
+			memset(bins, 0, sizeof(*bins));
+			return result;
+		}
+
+		/* Now bins->used is zero again. */
+
+		/*
+		 * Don't bother processing the last pile: when piles 0..N-1 are all in place, then
+		 * pile N must also be in place.
+		 */
+		end = task.last_key - bins->size[bins->last];
+		bins->size[bins->last] = 0;
+
+		for (fence = task.first_key; fence <= end; ) {
+			u8 bin;
+			sort_key_t key = *fence;
+
+			/*
+			 * The radix byte of the key tells us which pile it belongs in. Swap it for
+			 * an unprocessed item just below that pile, and repeat.
+			 */
+			while (--pile[bin = key[task.offset]] > fence)
+				swap_keys(pile[bin], &key);
+
+			/*
+			 * The pile reached the fence. Put the key at the bottom of that pile,
+			 * completing it, and advance the fence to the next pile.
+			 */
+			*fence = key;
+			fence += bins->size[bin];
+			bins->size[bin] = 0;
+		}
+
+		/* Now bins->size[] is all zero again. */
+
+		/*
+		 * When the number of keys in a task gets small enough, it is faster to use an
+		 * insertion sort than to keep subdividing into tiny piles.
+		 */
+		while (--insertion_task_list >= sorter->insertion_list)
+			insertion_sort(*insertion_task_list);
+	}
+
+	return UDS_SUCCESS;
+}