blob 5f069214d9060da8ceb001e6c037607c1d0096bf
1 #include "cache.h"
2 #include "sha1-lookup.h"
4 static uint32_t take2(const unsigned char *sha1)
6 return ((sha1[0] << 8) | sha1[1]);
9 /*
10 * Conventional binary search loop looks like this:
12 * do {
13 * int mi = (lo + hi) / 2;
14 * int cmp = "entry pointed at by mi" minus "target";
15 * if (!cmp)
16 * return (mi is the wanted one)
17 * if (cmp > 0)
18 * hi = mi; "mi is larger than target"
19 * else
20 * lo = mi+1; "mi is smaller than target"
21 * } while (lo < hi);
23 * The invariants are:
25 * - When entering the loop, lo points at a slot that is never
26 * above the target (it could be at the target), hi points at a
27 * slot that is guaranteed to be above the target (it can never
28 * be at the target).
30 * - We find a point 'mi' between lo and hi (mi could be the same
31 * as lo, but never can be the same as hi), and check if it hits
32 * the target. There are three cases:
34 * - if it is a hit, we are happy.
36 * - if it is strictly higher than the target, we update hi with
37 * it.
39 * - if it is strictly lower than the target, we update lo to be
40 * one slot after it, because we allow lo to be at the target.
42 * When choosing 'mi', we do not have to take the "middle" but
43 * anywhere in between lo and hi, as long as lo <= mi < hi is
44 * satisfied. When we somehow know that the distance between the
45 * target and lo is much shorter than the target and hi, we could
46 * pick mi that is much closer to lo than the midway.
49 * The table should contain "nr" elements.
50 * The sha1 of element i (between 0 and nr - 1) should be returned
51 * by "fn(i, table)".
53 int sha1_pos(const unsigned char *sha1, void *table, size_t nr,
54 sha1_access_fn fn)
56 size_t hi = nr;
57 size_t lo = 0;
58 size_t mi = 0;
60 if (!nr)
61 return -1;
63 if (nr != 1) {
64 size_t lov, hiv, miv, ofs;
66 for (ofs = 0; ofs < 18; ofs += 2) {
67 lov = take2(fn(0, table) + ofs);
68 hiv = take2(fn(nr - 1, table) + ofs);
69 miv = take2(sha1 + ofs);
70 if (miv < lov)
71 return -1;
72 if (hiv < miv)
73 return -1 - nr;
74 if (lov != hiv) {
76 * At this point miv could be equal
77 * to hiv (but sha1 could still be higher);
78 * the invariant of (mi < hi) should be
79 * kept.
81 mi = (nr - 1) * (miv - lov) / (hiv - lov);
82 if (lo <= mi && mi < hi)
83 break;
84 die("BUG: assertion failed in binary search");
89 do {
90 int cmp;
91 cmp = hashcmp(fn(mi, table), sha1);
92 if (!cmp)
93 return mi;
94 if (cmp > 0)
95 hi = mi;
96 else
97 lo = mi + 1;
98 mi = (hi + lo) / 2;
99 } while (lo < hi);
100 return -lo-1;
104 * Conventional binary search loop looks like this:
106 * unsigned lo, hi;
107 * do {
108 * unsigned mi = (lo + hi) / 2;
109 * int cmp = "entry pointed at by mi" minus "target";
110 * if (!cmp)
111 * return (mi is the wanted one)
112 * if (cmp > 0)
113 * hi = mi; "mi is larger than target"
114 * else
115 * lo = mi+1; "mi is smaller than target"
116 * } while (lo < hi);
118 * The invariants are:
120 * - When entering the loop, lo points at a slot that is never
121 * above the target (it could be at the target), hi points at a
122 * slot that is guaranteed to be above the target (it can never
123 * be at the target).
125 * - We find a point 'mi' between lo and hi (mi could be the same
126 * as lo, but never can be as same as hi), and check if it hits
127 * the target. There are three cases:
129 * - if it is a hit, we are happy.
131 * - if it is strictly higher than the target, we set it to hi,
132 * and repeat the search.
134 * - if it is strictly lower than the target, we update lo to
135 * one slot after it, because we allow lo to be at the target.
137 * If the loop exits, there is no matching entry.
139 * When choosing 'mi', we do not have to take the "middle" but
140 * anywhere in between lo and hi, as long as lo <= mi < hi is
141 * satisfied. When we somehow know that the distance between the
142 * target and lo is much shorter than the target and hi, we could
143 * pick mi that is much closer to lo than the midway.
145 * Now, we can take advantage of the fact that SHA-1 is a good hash
146 * function, and as long as there are enough entries in the table, we
147 * can expect uniform distribution. An entry that begins with for
148 * example "deadbeef..." is much likely to appear much later than in
149 * the midway of the table. It can reasonably be expected to be near
150 * 87% (222/256) from the top of the table.
152 * However, we do not want to pick "mi" too precisely. If the entry at
153 * the 87% in the above example turns out to be higher than the target
154 * we are looking for, we would end up narrowing the search space down
155 * only by 13%, instead of 50% we would get if we did a simple binary
156 * search. So we would want to hedge our bets by being less aggressive.
158 * The table at "table" holds at least "nr" entries of "elem_size"
159 * bytes each. Each entry has the SHA-1 key at "key_offset". The
160 * table is sorted by the SHA-1 key of the entries. The caller wants
161 * to find the entry with "key", and knows that the entry at "lo" is
162 * not higher than the entry it is looking for, and that the entry at
163 * "hi" is higher than the entry it is looking for.
165 int sha1_entry_pos(const void *table,
166 size_t elem_size,
167 size_t key_offset,
168 unsigned lo, unsigned hi, unsigned nr,
169 const unsigned char *key)
171 const unsigned char *base = table;
172 const unsigned char *hi_key, *lo_key;
173 unsigned ofs_0;
174 static int debug_lookup = -1;
176 if (debug_lookup < 0)
177 debug_lookup = !!getenv("GIT_DEBUG_LOOKUP");
179 if (!nr || lo >= hi)
180 return -1;
182 if (nr == hi)
183 hi_key = NULL;
184 else
185 hi_key = base + elem_size * hi + key_offset;
186 lo_key = base + elem_size * lo + key_offset;
188 ofs_0 = 0;
189 do {
190 int cmp;
191 unsigned ofs, mi, range;
192 unsigned lov, hiv, kyv;
193 const unsigned char *mi_key;
195 range = hi - lo;
196 if (hi_key) {
197 for (ofs = ofs_0; ofs < 20; ofs++)
198 if (lo_key[ofs] != hi_key[ofs])
199 break;
200 ofs_0 = ofs;
202 * byte 0 thru (ofs-1) are the same between
203 * lo and hi; ofs is the first byte that is
204 * different.
206 * If ofs==20, then no bytes are different,
207 * meaning we have entries with duplicate
208 * keys. We know that we are in a solid run
209 * of this entry (because the entries are
210 * sorted, and our lo and hi are the same,
211 * there can be nothing but this single key
212 * in between). So we can stop the search.
213 * Either one of these entries is it (and
214 * we do not care which), or we do not have
215 * it.
217 * Furthermore, we know that one of our
218 * endpoints must be the edge of the run of
219 * duplicates. For example, given this
220 * sequence:
222 * idx 0 1 2 3 4 5
223 * key A C C C C D
225 * If we are searching for "B", we might
226 * hit the duplicate run at lo=1, hi=3
227 * (e.g., by first mi=3, then mi=0). But we
228 * can never have lo > 1, because B < C.
229 * That is, if our key is less than the
230 * run, we know that "lo" is the edge, but
231 * we can say nothing of "hi". Similarly,
232 * if our key is greater than the run, we
233 * know that "hi" is the edge, but we can
234 * say nothing of "lo".
236 * Therefore if we do not find it, we also
237 * know where it would go if it did exist:
238 * just on the far side of the edge that we
239 * know about.
241 if (ofs == 20) {
242 mi = lo;
243 mi_key = base + elem_size * mi + key_offset;
244 cmp = memcmp(mi_key, key, 20);
245 if (!cmp)
246 return mi;
247 if (cmp < 0)
248 return -1 - hi;
249 else
250 return -1 - lo;
253 hiv = hi_key[ofs_0];
254 if (ofs_0 < 19)
255 hiv = (hiv << 8) | hi_key[ofs_0+1];
256 } else {
257 hiv = 256;
258 if (ofs_0 < 19)
259 hiv <<= 8;
261 lov = lo_key[ofs_0];
262 kyv = key[ofs_0];
263 if (ofs_0 < 19) {
264 lov = (lov << 8) | lo_key[ofs_0+1];
265 kyv = (kyv << 8) | key[ofs_0+1];
267 assert(lov < hiv);
269 if (kyv < lov)
270 return -1 - lo;
271 if (hiv < kyv)
272 return -1 - hi;
275 * Even if we know the target is much closer to 'hi'
276 * than 'lo', if we pick too precisely and overshoot
277 * (e.g. when we know 'mi' is closer to 'hi' than to
278 * 'lo', pick 'mi' that is higher than the target), we
279 * end up narrowing the search space by a smaller
280 * amount (i.e. the distance between 'mi' and 'hi')
281 * than what we would have (i.e. about half of 'lo'
282 * and 'hi'). Hedge our bets to pick 'mi' less
283 * aggressively, i.e. make 'mi' a bit closer to the
284 * middle than we would otherwise pick.
286 kyv = (kyv * 6 + lov + hiv) / 8;
287 if (lov < hiv - 1) {
288 if (kyv == lov)
289 kyv++;
290 else if (kyv == hiv)
291 kyv--;
293 mi = (range - 1) * (kyv - lov) / (hiv - lov) + lo;
295 if (debug_lookup) {
296 printf("lo %u hi %u rg %u mi %u ", lo, hi, range, mi);
297 printf("ofs %u lov %x, hiv %x, kyv %x\n",
298 ofs_0, lov, hiv, kyv);
300 if (!(lo <= mi && mi < hi))
301 die("assertion failure lo %u mi %u hi %u %s",
302 lo, mi, hi, sha1_to_hex(key));
304 mi_key = base + elem_size * mi + key_offset;
305 cmp = memcmp(mi_key + ofs_0, key + ofs_0, 20 - ofs_0);
306 if (!cmp)
307 return mi;
308 if (cmp > 0) {
309 hi = mi;
310 hi_key = mi_key;
311 } else {
312 lo = mi + 1;
313 lo_key = mi_key + elem_size;
315 } while (lo < hi);
316 return -lo-1;