Lab0 lib/list_sort.c 學習筆記

__attribute__((nonnull(2,3,4)))
static struct list_head *merge(void *priv, list_cmp_func_t cmp,
				struct list_head *a, struct list_head *b)
{
    ...
}

struct list_head *head, **tail = &head;

for (;;) {
		/* if equal, take 'a' -- important for sort stability */
		if (cmp(priv, a, b) <= 0) {
			*tail = a;
			tail = &a->next;
			a = a->next;
			if (!a) {
				*tail = b;
				break;
			}
		} else {
			*tail = b;
			tail = &b->next;
			b = b->next;
			if (!b) {
				*tail = a;
				break;
			}
		}
	}

__attribute__((nonnull(2,3,4,5)))
static void merge_final(void *priv, list_cmp_func_t cmp, struct list_head *head,
			struct list_head *a, struct list_head *b)

	/* Finish linking remainder of list b on to tail */
	tail->next = b;
	do {
		/*
		 * If the merge is highly unbalanced (e.g. the input is
		 * already sorted), this loop may run many iterations.
		 * Continue callbacks to the client even though no
		 * element comparison is needed, so the client's cmp()
		 * routine can invoke cond_resched() periodically.
		 */
		if (unlikely(!++count))
			cmp(priv, b, b);
		b->prev = tail;
		tail = b;
		b = b->next;
	} while (b);

	/* And the final links to make a circular doubly-linked list */
	tail->next = head;
	head->prev = tail;

# define likely(x)	__builtin_expect(!!(x), 1)
# define unlikely(x)	__builtin_expect(!!(x), 0)

 * This mergesort is as eager as possible while always performing at least
 * 2:1 balanced merges.  Given two pending sublists of size 2^k, they are
 * merged to a size-2^(k+1) list as soon as we have 2^k following elements.
 *
 * Thus, it will avoid cache thrashing as long as 3*2^k elements can
 * fit into the cache.  Not quite as good as a fully-eager bottom-up
 * mergesort, but it does use 0.2*n fewer comparisons, so is faster in
 * the common case that everything fits into L1.

 * The merging is controlled by "count", the number of elements in the
 * pending lists.  This is beautifully simple code, but rather subtle.
 *
 * Each time we increment "count", we set one bit (bit k) and clear
 * bits k-1 .. 0.  Each time this happens (except the very first time
 * for each bit, when count increments to 2^k), we merge two lists of
 * size 2^k into one list of size 2^(k+1).

 * This merge happens exactly when the count reaches an odd multiple of
 * 2^k, which is when we have 2^k elements pending in smaller lists,
 * so it's safe to merge away two lists of size 2^k.
 *
 * After this happens twice, we have created two lists of size 2^(k+1),
 * which will be merged into a list of size 2^(k+2) before we create
 * a third list of size 2^(k+1), so there are never more than two pending.

 * The number of pending lists of size 2^k is determined by the
 * state of bit k of "count" plus two extra pieces of information:
 *
 * - The state of bit k-1 (when k == 0, consider bit -1 always set), and
 * - Whether the higher-order bits are zero or non-zero (i.e.
 *   is count >= 2^(k+1)).

struct list_head *list = head->next, *pending = NULL;
size_t count = 0;	/* Count of pending */

if (list == head->prev)	/* Zero or one elements */
	return;

/* Convert to a null-terminated singly-linked list. */
head->prev->next = NULL;

size_t bits;
struct list_head **tail = &pending;

/* Find the least-significant clear bit in count */
for (bits = count; bits & 1; bits >>= 1)
	tail = &(*tail)->prev;

/* Do the indicated merge */
if (likely(bits)) {
	struct list_head *a = *tail, *b = a->prev;

	a = merge(priv, cmp, b, a);
    /* Install the merged result in place of the inputs */
	a->prev = b->prev;
	*tail = a;
}

/* Move one element from input list to pending */
list->prev = pending;
pending = list;
list = list->next;
pending->next = NULL;
count++;

/* End of input; merge together all the pending lists. */
list = pending;
pending = pending->prev;
for (;;) {
	struct list_head *next = pending->prev;

	if (!next)
		break;
	list = merge(priv, cmp, pending, list);
	pending = next;
}
/* The final merge, rebuilding prev links */
merge_final(priv, cmp, head, pending, list);