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bdiff: don't check border condition in loop...
Gregory Szorc -
r30308:d500ddae default
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@@ -1,296 +1,298 b''
1 1 /*
2 2 bdiff.c - efficient binary diff extension for Mercurial
3 3
4 4 Copyright 2005, 2006 Matt Mackall <mpm@selenic.com>
5 5
6 6 This software may be used and distributed according to the terms of
7 7 the GNU General Public License, incorporated herein by reference.
8 8
9 9 Based roughly on Python difflib
10 10 */
11 11
12 12 #include <stdlib.h>
13 13 #include <string.h>
14 14 #include <limits.h>
15 15
16 16 #include "compat.h"
17 17 #include "bitmanipulation.h"
18 18 #include "bdiff.h"
19 19
20 20 struct pos {
21 21 int pos, len;
22 22 };
23 23
24 24 int bdiff_splitlines(const char *a, ssize_t len, struct bdiff_line **lr)
25 25 {
26 26 unsigned hash;
27 27 int i;
28 28 const char *p, *b = a;
29 29 const char * const plast = a + len - 1;
30 30 struct bdiff_line *l;
31 31
32 32 /* count the lines */
33 33 i = 1; /* extra line for sentinel */
34 for (p = a; p < a + len; p++)
35 if (*p == '\n' || p == plast)
34 for (p = a; p < plast; p++)
35 if (*p == '\n')
36 36 i++;
37 if (p == plast)
38 i++;
37 39
38 40 *lr = l = (struct bdiff_line *)malloc(sizeof(struct bdiff_line) * i);
39 41 if (!l)
40 42 return -1;
41 43
42 44 /* build the line array and calculate hashes */
43 45 hash = 0;
44 46 for (p = a; p < a + len; p++) {
45 47 /* Leonid Yuriev's hash */
46 48 hash = (hash * 1664525) + (unsigned char)*p + 1013904223;
47 49
48 50 if (*p == '\n' || p == plast) {
49 51 l->hash = hash;
50 52 hash = 0;
51 53 l->len = p - b + 1;
52 54 l->l = b;
53 55 l->n = INT_MAX;
54 56 l++;
55 57 b = p + 1;
56 58 }
57 59 }
58 60
59 61 /* set up a sentinel */
60 62 l->hash = 0;
61 63 l->len = 0;
62 64 l->l = a + len;
63 65 return i - 1;
64 66 }
65 67
66 68 static inline int cmp(struct bdiff_line *a, struct bdiff_line *b)
67 69 {
68 70 return a->hash != b->hash || a->len != b->len || memcmp(a->l, b->l, a->len);
69 71 }
70 72
71 73 static int equatelines(struct bdiff_line *a, int an, struct bdiff_line *b,
72 74 int bn)
73 75 {
74 76 int i, j, buckets = 1, t, scale;
75 77 struct pos *h = NULL;
76 78
77 79 /* build a hash table of the next highest power of 2 */
78 80 while (buckets < bn + 1)
79 81 buckets *= 2;
80 82
81 83 /* try to allocate a large hash table to avoid collisions */
82 84 for (scale = 4; scale; scale /= 2) {
83 85 h = (struct pos *)malloc(scale * buckets * sizeof(struct pos));
84 86 if (h)
85 87 break;
86 88 }
87 89
88 90 if (!h)
89 91 return 0;
90 92
91 93 buckets = buckets * scale - 1;
92 94
93 95 /* clear the hash table */
94 96 for (i = 0; i <= buckets; i++) {
95 97 h[i].pos = -1;
96 98 h[i].len = 0;
97 99 }
98 100
99 101 /* add lines to the hash table chains */
100 102 for (i = 0; i < bn; i++) {
101 103 /* find the equivalence class */
102 104 for (j = b[i].hash & buckets; h[j].pos != -1;
103 105 j = (j + 1) & buckets)
104 106 if (!cmp(b + i, b + h[j].pos))
105 107 break;
106 108
107 109 /* add to the head of the equivalence class */
108 110 b[i].n = h[j].pos;
109 111 b[i].e = j;
110 112 h[j].pos = i;
111 113 h[j].len++; /* keep track of popularity */
112 114 }
113 115
114 116 /* compute popularity threshold */
115 117 t = (bn >= 31000) ? bn / 1000 : 1000000 / (bn + 1);
116 118
117 119 /* match items in a to their equivalence class in b */
118 120 for (i = 0; i < an; i++) {
119 121 /* find the equivalence class */
120 122 for (j = a[i].hash & buckets; h[j].pos != -1;
121 123 j = (j + 1) & buckets)
122 124 if (!cmp(a + i, b + h[j].pos))
123 125 break;
124 126
125 127 a[i].e = j; /* use equivalence class for quick compare */
126 128 if (h[j].len <= t)
127 129 a[i].n = h[j].pos; /* point to head of match list */
128 130 else
129 131 a[i].n = -1; /* too popular */
130 132 }
131 133
132 134 /* discard hash tables */
133 135 free(h);
134 136 return 1;
135 137 }
136 138
137 139 static int longest_match(struct bdiff_line *a, struct bdiff_line *b,
138 140 struct pos *pos,
139 141 int a1, int a2, int b1, int b2, int *omi, int *omj)
140 142 {
141 143 int mi = a1, mj = b1, mk = 0, i, j, k, half;
142 144
143 145 /* window our search on large regions to better bound
144 146 worst-case performance. by choosing a window at the end, we
145 147 reduce skipping overhead on the b chains. */
146 148 if (a2 - a1 > 30000)
147 149 a1 = a2 - 30000;
148 150
149 151 half = (a1 + a2) / 2;
150 152
151 153 for (i = a1; i < a2; i++) {
152 154 /* skip all lines in b after the current block */
153 155 for (j = a[i].n; j >= b2; j = b[j].n)
154 156 ;
155 157
156 158 /* loop through all lines match a[i] in b */
157 159 for (; j >= b1; j = b[j].n) {
158 160 /* does this extend an earlier match? */
159 161 for (k = 1; j - k >= b1 && i - k >= a1; k++) {
160 162 /* reached an earlier match? */
161 163 if (pos[j - k].pos == i - k) {
162 164 k += pos[j - k].len;
163 165 break;
164 166 }
165 167 /* previous line mismatch? */
166 168 if (a[i - k].e != b[j - k].e)
167 169 break;
168 170 }
169 171
170 172 pos[j].pos = i;
171 173 pos[j].len = k;
172 174
173 175 /* best match so far? we prefer matches closer
174 176 to the middle to balance recursion */
175 177 if (k > mk || (k == mk && (i <= mi || i < half))) {
176 178 mi = i;
177 179 mj = j;
178 180 mk = k;
179 181 }
180 182 }
181 183 }
182 184
183 185 if (mk) {
184 186 mi = mi - mk + 1;
185 187 mj = mj - mk + 1;
186 188 }
187 189
188 190 /* expand match to include subsequent popular lines */
189 191 while (mi + mk < a2 && mj + mk < b2 &&
190 192 a[mi + mk].e == b[mj + mk].e)
191 193 mk++;
192 194
193 195 *omi = mi;
194 196 *omj = mj;
195 197
196 198 return mk;
197 199 }
198 200
199 201 static struct bdiff_hunk *recurse(struct bdiff_line *a, struct bdiff_line *b,
200 202 struct pos *pos,
201 203 int a1, int a2, int b1, int b2, struct bdiff_hunk *l)
202 204 {
203 205 int i, j, k;
204 206
205 207 while (1) {
206 208 /* find the longest match in this chunk */
207 209 k = longest_match(a, b, pos, a1, a2, b1, b2, &i, &j);
208 210 if (!k)
209 211 return l;
210 212
211 213 /* and recurse on the remaining chunks on either side */
212 214 l = recurse(a, b, pos, a1, i, b1, j, l);
213 215 if (!l)
214 216 return NULL;
215 217
216 218 l->next = (struct bdiff_hunk *)malloc(sizeof(struct bdiff_hunk));
217 219 if (!l->next)
218 220 return NULL;
219 221
220 222 l = l->next;
221 223 l->a1 = i;
222 224 l->a2 = i + k;
223 225 l->b1 = j;
224 226 l->b2 = j + k;
225 227 l->next = NULL;
226 228
227 229 /* tail-recursion didn't happen, so do equivalent iteration */
228 230 a1 = i + k;
229 231 b1 = j + k;
230 232 }
231 233 }
232 234
233 235 int bdiff_diff(struct bdiff_line *a, int an, struct bdiff_line *b,
234 236 int bn, struct bdiff_hunk *base)
235 237 {
236 238 struct bdiff_hunk *curr;
237 239 struct pos *pos;
238 240 int t, count = 0;
239 241
240 242 /* allocate and fill arrays */
241 243 t = equatelines(a, an, b, bn);
242 244 pos = (struct pos *)calloc(bn ? bn : 1, sizeof(struct pos));
243 245
244 246 if (pos && t) {
245 247 /* generate the matching block list */
246 248
247 249 curr = recurse(a, b, pos, 0, an, 0, bn, base);
248 250 if (!curr)
249 251 return -1;
250 252
251 253 /* sentinel end hunk */
252 254 curr->next = (struct bdiff_hunk *)malloc(sizeof(struct bdiff_hunk));
253 255 if (!curr->next)
254 256 return -1;
255 257 curr = curr->next;
256 258 curr->a1 = curr->a2 = an;
257 259 curr->b1 = curr->b2 = bn;
258 260 curr->next = NULL;
259 261 }
260 262
261 263 free(pos);
262 264
263 265 /* normalize the hunk list, try to push each hunk towards the end */
264 266 for (curr = base->next; curr; curr = curr->next) {
265 267 struct bdiff_hunk *next = curr->next;
266 268
267 269 if (!next)
268 270 break;
269 271
270 272 if (curr->a2 == next->a1 || curr->b2 == next->b1)
271 273 while (curr->a2 < an && curr->b2 < bn
272 274 && next->a1 < next->a2
273 275 && next->b1 < next->b2
274 276 && !cmp(a + curr->a2, b + curr->b2)) {
275 277 curr->a2++;
276 278 next->a1++;
277 279 curr->b2++;
278 280 next->b1++;
279 281 }
280 282 }
281 283
282 284 for (curr = base->next; curr; curr = curr->next)
283 285 count++;
284 286 return count;
285 287 }
286 288
287 289 void bdiff_freehunks(struct bdiff_hunk *l)
288 290 {
289 291 struct bdiff_hunk *n;
290 292 for (; l; l = n) {
291 293 n = l->next;
292 294 free(l);
293 295 }
294 296 }
295 297
296 298
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