##// END OF EJS Templates
snapshot: search for unrelated but reusable full-snapshot...
snapshot: search for unrelated but reusable full-snapshot # New Strategy Step: Reusing Snapshot Outside Of Parents' Chain. If no suitable bases were found in the parent's chains, see if we could reuse a full snapshot not directly related to the current revision. Such search can be expensive, so we only search for snapshots appended to the revlog *after* the bases used by the parents of the current revision (the one we just tested). We assume the parent's bases were created because the previous snapshots were unsuitable, so there are low odds they would be useful now. This search gives a chance to reuse a delta chain unrelated to the current revision. Without this re-use, topological branches would keep reopening new full chains. Creating more and more snapshots as the repository grow. In repositories with many topological branches, the lack of delta reuse can create too many snapshots reducing overall compression to nothing. This results in a very large repository and other usability issues. For now, we still focus on creating level-1 snapshots. However, this principle will play a large part in how we avoid snapshot explosion once we have more snapshot levels. # Effects On The Test Repository In the test repository we created, we can see the beneficial effect of such reuse. We need very few level-0 snapshots and the overall revlog size has decreased. The `hg debugrevlog` call, show a "lvl-2" snapshot. It comes from the existing delta logic using the `prev` revision (revlog's tip) as the base. In this specific case, it turns out the tip was a level-1 snapshot. This is a coincidence that can be ignored. Finding and testing against all these unrelated snapshots can have a performance impact at write time. We currently focus on building good deltas chain we build. Performance concern will be dealt with later in another series.

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bdiff.c
321 lines | 6.9 KiB | text/x-c | CLexer
/*
bdiff.c - efficient binary diff extension for Mercurial
Copyright 2005, 2006 Matt Mackall <mpm@selenic.com>
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);
}
}