sha1-lookup.con commit rebase: fix fork-point with zero arguments (bb3f458)
   1#include "cache.h"
   2#include "sha1-lookup.h"
   3
   4static uint32_t take2(const unsigned char *sha1)
   5{
   6        return ((sha1[0] << 8) | sha1[1]);
   7}
   8
   9/*
  10 * Conventional binary search loop looks like this:
  11 *
  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);
  22 *
  23 * The invariants are:
  24 *
  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).
  29 *
  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:
  33 *
  34 *    - if it is a hit, we are happy.
  35 *
  36 *    - if it is strictly higher than the target, we update hi with
  37 *      it.
  38 *
  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.
  41 *
  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.
  47 */
  48/*
  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)".
  52 */
  53int sha1_pos(const unsigned char *sha1, void *table, size_t nr,
  54             sha1_access_fn fn)
  55{
  56        size_t hi = nr;
  57        size_t lo = 0;
  58        size_t mi = 0;
  59
  60        if (!nr)
  61                return -1;
  62
  63        if (nr != 1) {
  64                size_t lov, hiv, miv, ofs;
  65
  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) {
  75                                /*
  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.
  80                                 */
  81                                mi = (nr - 1) * (miv - lov) / (hiv - lov);
  82                                if (lo <= mi && mi < hi)
  83                                        break;
  84                                die("BUG: assertion failed in binary search");
  85                        }
  86                }
  87                if (18 <= ofs)
  88                        die("cannot happen -- lo and hi are identical");
  89        }
  90
  91        do {
  92                int cmp;
  93                cmp = hashcmp(fn(mi, table), sha1);
  94                if (!cmp)
  95                        return mi;
  96                if (cmp > 0)
  97                        hi = mi;
  98                else
  99                        lo = mi + 1;
 100                mi = (hi + lo) / 2;
 101        } while (lo < hi);
 102        return -lo-1;
 103}
 104
 105/*
 106 * Conventional binary search loop looks like this:
 107 *
 108 *      unsigned lo, hi;
 109 *      do {
 110 *              unsigned mi = (lo + hi) / 2;
 111 *              int cmp = "entry pointed at by mi" minus "target";
 112 *              if (!cmp)
 113 *                      return (mi is the wanted one)
 114 *              if (cmp > 0)
 115 *                      hi = mi; "mi is larger than target"
 116 *              else
 117 *                      lo = mi+1; "mi is smaller than target"
 118 *      } while (lo < hi);
 119 *
 120 * The invariants are:
 121 *
 122 * - When entering the loop, lo points at a slot that is never
 123 *   above the target (it could be at the target), hi points at a
 124 *   slot that is guaranteed to be above the target (it can never
 125 *   be at the target).
 126 *
 127 * - We find a point 'mi' between lo and hi (mi could be the same
 128 *   as lo, but never can be as same as hi), and check if it hits
 129 *   the target.  There are three cases:
 130 *
 131 *    - if it is a hit, we are happy.
 132 *
 133 *    - if it is strictly higher than the target, we set it to hi,
 134 *      and repeat the search.
 135 *
 136 *    - if it is strictly lower than the target, we update lo to
 137 *      one slot after it, because we allow lo to be at the target.
 138 *
 139 *   If the loop exits, there is no matching entry.
 140 *
 141 * When choosing 'mi', we do not have to take the "middle" but
 142 * anywhere in between lo and hi, as long as lo <= mi < hi is
 143 * satisfied.  When we somehow know that the distance between the
 144 * target and lo is much shorter than the target and hi, we could
 145 * pick mi that is much closer to lo than the midway.
 146 *
 147 * Now, we can take advantage of the fact that SHA-1 is a good hash
 148 * function, and as long as there are enough entries in the table, we
 149 * can expect uniform distribution.  An entry that begins with for
 150 * example "deadbeef..." is much likely to appear much later than in
 151 * the midway of the table.  It can reasonably be expected to be near
 152 * 87% (222/256) from the top of the table.
 153 *
 154 * However, we do not want to pick "mi" too precisely.  If the entry at
 155 * the 87% in the above example turns out to be higher than the target
 156 * we are looking for, we would end up narrowing the search space down
 157 * only by 13%, instead of 50% we would get if we did a simple binary
 158 * search.  So we would want to hedge our bets by being less aggressive.
 159 *
 160 * The table at "table" holds at least "nr" entries of "elem_size"
 161 * bytes each.  Each entry has the SHA-1 key at "key_offset".  The
 162 * table is sorted by the SHA-1 key of the entries.  The caller wants
 163 * to find the entry with "key", and knows that the entry at "lo" is
 164 * not higher than the entry it is looking for, and that the entry at
 165 * "hi" is higher than the entry it is looking for.
 166 */
 167int sha1_entry_pos(const void *table,
 168                   size_t elem_size,
 169                   size_t key_offset,
 170                   unsigned lo, unsigned hi, unsigned nr,
 171                   const unsigned char *key)
 172{
 173        const unsigned char *base = table;
 174        const unsigned char *hi_key, *lo_key;
 175        unsigned ofs_0;
 176        static int debug_lookup = -1;
 177
 178        if (debug_lookup < 0)
 179                debug_lookup = !!getenv("GIT_DEBUG_LOOKUP");
 180
 181        if (!nr || lo >= hi)
 182                return -1;
 183
 184        if (nr == hi)
 185                hi_key = NULL;
 186        else
 187                hi_key = base + elem_size * hi + key_offset;
 188        lo_key = base + elem_size * lo + key_offset;
 189
 190        ofs_0 = 0;
 191        do {
 192                int cmp;
 193                unsigned ofs, mi, range;
 194                unsigned lov, hiv, kyv;
 195                const unsigned char *mi_key;
 196
 197                range = hi - lo;
 198                if (hi_key) {
 199                        for (ofs = ofs_0; ofs < 20; ofs++)
 200                                if (lo_key[ofs] != hi_key[ofs])
 201                                        break;
 202                        ofs_0 = ofs;
 203                        /*
 204                         * byte 0 thru (ofs-1) are the same between
 205                         * lo and hi; ofs is the first byte that is
 206                         * different.
 207                         *
 208                         * If ofs==20, then no bytes are different,
 209                         * meaning we have entries with duplicate
 210                         * keys. We know that we are in a solid run
 211                         * of this entry (because the entries are
 212                         * sorted, and our lo and hi are the same,
 213                         * there can be nothing but this single key
 214                         * in between). So we can stop the search.
 215                         * Either one of these entries is it (and
 216                         * we do not care which), or we do not have
 217                         * it.
 218                         *
 219                         * Furthermore, we know that one of our
 220                         * endpoints must be the edge of the run of
 221                         * duplicates. For example, given this
 222                         * sequence:
 223                         *
 224                         *     idx 0 1 2 3 4 5
 225                         *     key A C C C C D
 226                         *
 227                         * If we are searching for "B", we might
 228                         * hit the duplicate run at lo=1, hi=3
 229                         * (e.g., by first mi=3, then mi=0). But we
 230                         * can never have lo > 1, because B < C.
 231                         * That is, if our key is less than the
 232                         * run, we know that "lo" is the edge, but
 233                         * we can say nothing of "hi". Similarly,
 234                         * if our key is greater than the run, we
 235                         * know that "hi" is the edge, but we can
 236                         * say nothing of "lo".
 237                         *
 238                         * Therefore if we do not find it, we also
 239                         * know where it would go if it did exist:
 240                         * just on the far side of the edge that we
 241                         * know about.
 242                         */
 243                        if (ofs == 20) {
 244                                mi = lo;
 245                                mi_key = base + elem_size * mi + key_offset;
 246                                cmp = memcmp(mi_key, key, 20);
 247                                if (!cmp)
 248                                        return mi;
 249                                if (cmp < 0)
 250                                        return -1 - hi;
 251                                else
 252                                        return -1 - lo;
 253                        }
 254
 255                        hiv = hi_key[ofs_0];
 256                        if (ofs_0 < 19)
 257                                hiv = (hiv << 8) | hi_key[ofs_0+1];
 258                } else {
 259                        hiv = 256;
 260                        if (ofs_0 < 19)
 261                                hiv <<= 8;
 262                }
 263                lov = lo_key[ofs_0];
 264                kyv = key[ofs_0];
 265                if (ofs_0 < 19) {
 266                        lov = (lov << 8) | lo_key[ofs_0+1];
 267                        kyv = (kyv << 8) | key[ofs_0+1];
 268                }
 269                assert(lov < hiv);
 270
 271                if (kyv < lov)
 272                        return -1 - lo;
 273                if (hiv < kyv)
 274                        return -1 - hi;
 275
 276                /*
 277                 * Even if we know the target is much closer to 'hi'
 278                 * than 'lo', if we pick too precisely and overshoot
 279                 * (e.g. when we know 'mi' is closer to 'hi' than to
 280                 * 'lo', pick 'mi' that is higher than the target), we
 281                 * end up narrowing the search space by a smaller
 282                 * amount (i.e. the distance between 'mi' and 'hi')
 283                 * than what we would have (i.e. about half of 'lo'
 284                 * and 'hi').  Hedge our bets to pick 'mi' less
 285                 * aggressively, i.e. make 'mi' a bit closer to the
 286                 * middle than we would otherwise pick.
 287                 */
 288                kyv = (kyv * 6 + lov + hiv) / 8;
 289                if (lov < hiv - 1) {
 290                        if (kyv == lov)
 291                                kyv++;
 292                        else if (kyv == hiv)
 293                                kyv--;
 294                }
 295                mi = (range - 1) * (kyv - lov) / (hiv - lov) + lo;
 296
 297                if (debug_lookup) {
 298                        printf("lo %u hi %u rg %u mi %u ", lo, hi, range, mi);
 299                        printf("ofs %u lov %x, hiv %x, kyv %x\n",
 300                               ofs_0, lov, hiv, kyv);
 301                }
 302                if (!(lo <= mi && mi < hi))
 303                        die("assertion failure lo %u mi %u hi %u %s",
 304                            lo, mi, hi, sha1_to_hex(key));
 305
 306                mi_key = base + elem_size * mi + key_offset;
 307                cmp = memcmp(mi_key + ofs_0, key + ofs_0, 20 - ofs_0);
 308                if (!cmp)
 309                        return mi;
 310                if (cmp > 0) {
 311                        hi = mi;
 312                        hi_key = mi_key;
 313                } else {
 314                        lo = mi + 1;
 315                        lo_key = mi_key + elem_size;
 316                }
 317        } while (lo < hi);
 318        return -lo-1;
 319}