view mercurial/bdiff.c @ 32697:19b9fc40cc51

revlog: skeleton support for version 2 revlogs

There are a number of improvements we want to make to revlogs
that will require a new version - version 2. It is unclear what the
full set of improvements will be or when we'll be done with them.
What I do know is that the process will likely take longer than a
single release, will require input from various stakeholders to
evaluate changes, and will have many contentious debates and
bikeshedding.

It is unrealistic to develop revlog version 2 up front: there
are just too many uncertainties that we won't know until things
are implemented and experiments are run. Some changes will also
be invasive and prone to bit rot, so sitting on dozens of patches
is not practical.

This commit introduces skeleton support for version 2 revlogs in
a way that is flexible and not bound by backwards compatibility
concerns.

An experimental repo requirement for denoting revlog v2 has been
added. The requirement string has a sub-version component to it.
This will allow us to declare multiple requirements in the course
of developing revlog v2. Whenever we change the in-development
revlog v2 format, we can tweak the string, creating a new
requirement and locking out old clients. This will allow us to
make as many backwards incompatible changes and experiments to
revlog v2 as we want. In other words, we can land code and make
meaningful progress towards revlog v2 while still maintaining
extreme format flexibility up until the point we freeze the
format and remove the experimental labels.

To enable the new repo requirement, you must supply an experimental
and undocumented config option. But not just any boolean flag
will do: you need to explicitly use a value that no sane person
should ever type. This is an additional guard against enabling
revlog v2 on an installation it shouldn't be enabled on. The
specific scenario I'm trying to prevent is say a user with a
4.4 client with a frozen format enabling the option but then
downgrading to 4.3 and accidentally creating repos with an
outdated and unsupported repo format. Requiring a "challenge"
string should prevent this.

Because the format is not yet finalized and I don't want to take
any chances, revlog v2's version is currently 0xDEAD. I figure
squatting on a value we're likely never to use as an actual revlog
version to mean "internal testing only" is acceptable. And
"dead" is easily recognized as something meaningful.

There is a bunch of cleanup that is needed before work on revlog
v2 begins in earnest. I plan on doing that work once this patch
is accepted and we're comfortable with the idea of starting down
this path.

/*
 bdiff.c - efficient binary diff extension for Mercurial

 Copyright 2005, 2006 Matt Mackall <[e-mail]>

 This software may be used and distributed according to the terms of
 the GNU General Public License, incorporated herein by reference.

 Based roughly on Python difflib
*/

#include <stdlib.h>
#include <string.h>
#include <limits.h>

#include "compat.h"
#include "bitmanipulation.h"
#include "bdiff.h"

/* Hash implementation from diffutils */
#define ROL(v, n) ((v) << (n) | (v) >> (sizeof(v) * CHAR_BIT - (n)))
#define HASH(h, c) ((c) + ROL(h ,7))

struct pos {
	int pos, len;
};

int bdiff_splitlines(const char *a, ssize_t len, struct bdiff_line **lr)
{
	unsigned hash;
	int i;
	const char *p, *b = a;
	const char * const plast = a + len - 1;
	struct bdiff_line *l;

	/* count the lines */
	i = 1; /* extra line for sentinel */
	for (p = a; p < plast; p++)
		if (*p == '\n')
			i++;
	if (p == plast)
		i++;

	*lr = l = (struct bdiff_line *)malloc(sizeof(struct bdiff_line) * i);
	if (!l)
		return -1;

	/* build the line array and calculate hashes */
	hash = 0;
	for (p = a; p < plast; p++) {
		hash = HASH(hash, *p);

		if (*p == '\n') {
			l->hash = hash;
			hash = 0;
			l->len = p - b + 1;
			l->l = b;
			l->n = INT_MAX;
			l++;
			b = p + 1;
		}
	}

	if (p == plast) {
		hash = HASH(hash, *p);
		l->hash = hash;
		l->len = p - b + 1;
		l->l = b;
		l->n = INT_MAX;
		l++;
	}

	/* set up a sentinel */
	l->hash = 0;
	l->len = 0;
	l->l = a + len;
	return i - 1;
}

static inline int cmp(struct bdiff_line *a, struct bdiff_line *b)
{
	return a->hash != b->hash || a->len != b->len || memcmp(a->l, b->l, a->len);
}

static int equatelines(struct bdiff_line *a, int an, struct bdiff_line *b,
	int bn)
{
	int i, j, buckets = 1, t, scale;
	struct pos *h = NULL;

	/* build a hash table of the next highest power of 2 */
	while (buckets < bn + 1)
		buckets *= 2;

	/* try to allocate a large hash table to avoid collisions */
	for (scale = 4; scale; scale /= 2) {
		h = (struct pos *)malloc(scale * buckets * sizeof(struct pos));
		if (h)
			break;
	}

	if (!h)
		return 0;

	buckets = buckets * scale - 1;

	/* clear the hash table */
	for (i = 0; i <= buckets; i++) {
		h[i].pos = -1;
		h[i].len = 0;
	}

	/* add lines to the hash table chains */
	for (i = 0; i < bn; i++) {
		/* find the equivalence class */
		for (j = b[i].hash & buckets; h[j].pos != -1;
		     j = (j + 1) & buckets)
			if (!cmp(b + i, b + h[j].pos))
				break;

		/* add to the head of the equivalence class */
		b[i].n = h[j].pos;
		b[i].e = j;
		h[j].pos = i;
		h[j].len++; /* keep track of popularity */
	}

	/* compute popularity threshold */
	t = (bn >= 31000) ? bn / 1000 : 1000000 / (bn + 1);

	/* match items in a to their equivalence class in b */
	for (i = 0; i < an; i++) {
		/* find the equivalence class */
		for (j = a[i].hash & buckets; h[j].pos != -1;
		     j = (j + 1) & buckets)
			if (!cmp(a + i, b + h[j].pos))
				break;

		a[i].e = j; /* use equivalence class for quick compare */
		if (h[j].len <= t)
			a[i].n = h[j].pos; /* point to head of match list */
		else
			a[i].n = -1; /* too popular */
	}

	/* discard hash tables */
	free(h);
	return 1;
}

static int longest_match(struct bdiff_line *a, struct bdiff_line *b,
			struct pos *pos,
			 int a1, int a2, int b1, int b2, int *omi, int *omj)
{
	int mi = a1, mj = b1, mk = 0, i, j, k, half, bhalf;

	/* window our search on large regions to better bound
	   worst-case performance. by choosing a window at the end, we
	   reduce skipping overhead on the b chains. */
	if (a2 - a1 > 30000)
		a1 = a2 - 30000;

	half = (a1 + a2 - 1) / 2;
	bhalf = (b1 + b2 - 1) / 2;

	for (i = a1; i < a2; i++) {
		/* skip all lines in b after the current block */
		for (j = a[i].n; j >= b2; j = b[j].n)
			;

		/* loop through all lines match a[i] in b */
		for (; j >= b1; j = b[j].n) {
			/* does this extend an earlier match? */
			for (k = 1; j - k >= b1 && i - k >= a1; k++) {
				/* reached an earlier match? */
				if (pos[j - k].pos == i - k) {
					k += pos[j - k].len;
					break;
				}
				/* previous line mismatch? */
				if (a[i - k].e != b[j - k].e)
					break;
			}

			pos[j].pos = i;
			pos[j].len = k;

			/* best match so far? we prefer matches closer
			   to the middle to balance recursion */
			if (k > mk) {
				/* a longer match */
				mi = i;
				mj = j;
				mk = k;
			} else if (k == mk) {
				if (i > mi && i <= half && j > b1) {
					/* same match but closer to half */
					mi = i;
					mj = j;
				} else if (i == mi && (mj > bhalf || i == a1)) {
					/* same i but best earlier j */
					mj = j;
				}
			}
		}
	}

	if (mk) {
		mi = mi - mk + 1;
		mj = mj - mk + 1;
	}

	/* expand match to include subsequent popular lines */
	while (mi + mk < a2 && mj + mk < b2 &&
	       a[mi + mk].e == b[mj + mk].e)
		mk++;

	*omi = mi;
	*omj = mj;

	return mk;
}

static struct bdiff_hunk *recurse(struct bdiff_line *a, struct bdiff_line *b,
				struct pos *pos,
			    int a1, int a2, int b1, int b2, struct bdiff_hunk *l)
{
	int i, j, k;

	while (1) {
		/* find the longest match in this chunk */
		k = longest_match(a, b, pos, a1, a2, b1, b2, &i, &j);
		if (!k)
			return l;

		/* and recurse on the remaining chunks on either side */
		l = recurse(a, b, pos, a1, i, b1, j, l);
		if (!l)
			return NULL;

		l->next = (struct bdiff_hunk *)malloc(sizeof(struct bdiff_hunk));
		if (!l->next)
			return NULL;

		l = l->next;
		l->a1 = i;
		l->a2 = i + k;
		l->b1 = j;
		l->b2 = j + k;
		l->next = NULL;

		/* tail-recursion didn't happen, so do equivalent iteration */
		a1 = i + k;
		b1 = j + k;
	}
}

int bdiff_diff(struct bdiff_line *a, int an, struct bdiff_line *b,
		int bn, struct bdiff_hunk *base)
{
	struct bdiff_hunk *curr;
	struct pos *pos;
	int t, count = 0;

	/* allocate and fill arrays */
	t = equatelines(a, an, b, bn);
	pos = (struct pos *)calloc(bn ? bn : 1, sizeof(struct pos));

	if (pos && t) {
		/* generate the matching block list */

		curr = recurse(a, b, pos, 0, an, 0, bn, base);
		if (!curr)
			return -1;

		/* sentinel end hunk */
		curr->next = (struct bdiff_hunk *)malloc(sizeof(struct bdiff_hunk));
		if (!curr->next)
			return -1;
		curr = curr->next;
		curr->a1 = curr->a2 = an;
		curr->b1 = curr->b2 = bn;
		curr->next = NULL;
	}

	free(pos);

	/* normalize the hunk list, try to push each hunk towards the end */
	for (curr = base->next; curr; curr = curr->next) {
		struct bdiff_hunk *next = curr->next;

		if (!next)
			break;

		if (curr->a2 == next->a1 || curr->b2 == next->b1)
			while (curr->a2 < an && curr->b2 < bn
			       && next->a1 < next->a2
			       && next->b1 < next->b2
			       && !cmp(a + curr->a2, b + curr->b2)) {
				curr->a2++;
				next->a1++;
				curr->b2++;
				next->b1++;
			}
	}

	for (curr = base->next; curr; curr = curr->next)
		count++;
	return count;
}

void bdiff_freehunks(struct bdiff_hunk *l)
{
	struct bdiff_hunk *n;
	for (; l; l = n) {
		n = l->next;
		free(l);
	}
}