mercurial/bdiff.c

perf: use storage API for resolving manifest node

lookup() isn't part of the storage API. And this code shouldn't
be accessing manifestlog._revlog directly for the modern code base.
So let's port it to the modern API.

Note that the previous code was busted for cases where we needed
to call lookup() because lookup() isn't exposed by manifestrevlog
any more.

This change is strictly BC breaking because we no longer support
resolving partial nodes. But it is a perf* command and I don't
think we should flag the change as such.

Differential Revision: https://phab.mercurial-scm.org/D4390

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

#include "bdiff.h"
#include "bitmanipulation.h"
#include "compat.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 *)calloc(i, sizeof(struct bdiff_line));
	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 *)calloc(buckets, scale * 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;
}

/* deallocate list of hunks; l may be NULL */
void bdiff_freehunks(struct bdiff_hunk *l)
{
	struct bdiff_hunk *n;
	for (; l; l = n) {
		n = l->next;
		free(l);
	}
}