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1 | 1 | /* |
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2 | 2 | bdiff.c - efficient binary diff extension for Mercurial |
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3 | 3 | |
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4 | 4 | Copyright 2005, 2006 Matt Mackall <mpm@selenic.com> |
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5 | 5 | |
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6 | 6 | This software may be used and distributed according to the terms of |
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7 | 7 | the GNU General Public License, incorporated herein by reference. |
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8 | 8 | |
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9 | 9 | Based roughly on Python difflib |
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10 | 10 | */ |
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11 | 11 | |
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12 | 12 | #include <stdlib.h> |
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13 | 13 | #include <string.h> |
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14 | 14 | #include <limits.h> |
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15 | 15 | |
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16 | 16 | #include "compat.h" |
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17 | 17 | #include "bitmanipulation.h" |
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18 | 18 | #include "bdiff.h" |
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19 | 19 | |
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20 | 20 | /* Hash implementation from diffutils */ |
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21 | 21 | #define ROL(v, n) ((v) << (n) | (v) >> (sizeof(v) * CHAR_BIT - (n))) |
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22 | 22 | #define HASH(h, c) ((c) + ROL(h ,7)) |
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23 | 23 | |
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24 | 24 | struct pos { |
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25 | 25 | int pos, len; |
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26 | 26 | }; |
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27 | 27 | |
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28 | 28 | int bdiff_splitlines(const char *a, ssize_t len, struct bdiff_line **lr) |
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29 | 29 | { |
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30 | 30 | unsigned hash; |
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31 | 31 | int i; |
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32 | 32 | const char *p, *b = a; |
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33 | 33 | const char * const plast = a + len - 1; |
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34 | 34 | struct bdiff_line *l; |
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35 | 35 | |
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36 | 36 | /* count the lines */ |
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37 | 37 | i = 1; /* extra line for sentinel */ |
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38 | 38 | for (p = a; p < plast; p++) |
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39 | 39 | if (*p == '\n') |
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40 | 40 | i++; |
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41 | 41 | if (p == plast) |
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42 | 42 | i++; |
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43 | 43 | |
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44 | 44 | *lr = l = (struct bdiff_line *)malloc(sizeof(struct bdiff_line) * i); |
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45 | 45 | if (!l) |
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46 | 46 | return -1; |
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47 | 47 | |
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48 | 48 | /* build the line array and calculate hashes */ |
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49 | 49 | hash = 0; |
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50 |
for (p = a; p < |
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50 | for (p = a; p < plast; p++) { | |
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51 | 51 | hash = HASH(hash, *p); |
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52 | 52 | |
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53 |
if (*p == '\n' |
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53 | if (*p == '\n') { | |
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54 | 54 | l->hash = hash; |
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55 | 55 | hash = 0; |
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56 | 56 | l->len = p - b + 1; |
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57 | 57 | l->l = b; |
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58 | 58 | l->n = INT_MAX; |
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59 | 59 | l++; |
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60 | 60 | b = p + 1; |
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61 | 61 | } |
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62 | 62 | } |
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63 | 63 | |
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64 | if (p == plast) { | |
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65 | hash = HASH(hash, *p); | |
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66 | l->hash = hash; | |
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67 | l->len = p - b + 1; | |
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68 | l->l = b; | |
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69 | l->n = INT_MAX; | |
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70 | l++; | |
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71 | } | |
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72 | ||
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64 | 73 | /* set up a sentinel */ |
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65 | 74 | l->hash = 0; |
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66 | 75 | l->len = 0; |
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67 | 76 | l->l = a + len; |
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68 | 77 | return i - 1; |
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69 | 78 | } |
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70 | 79 | |
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71 | 80 | static inline int cmp(struct bdiff_line *a, struct bdiff_line *b) |
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72 | 81 | { |
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73 | 82 | return a->hash != b->hash || a->len != b->len || memcmp(a->l, b->l, a->len); |
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74 | 83 | } |
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75 | 84 | |
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76 | 85 | static int equatelines(struct bdiff_line *a, int an, struct bdiff_line *b, |
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77 | 86 | int bn) |
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78 | 87 | { |
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79 | 88 | int i, j, buckets = 1, t, scale; |
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80 | 89 | struct pos *h = NULL; |
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81 | 90 | |
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82 | 91 | /* build a hash table of the next highest power of 2 */ |
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83 | 92 | while (buckets < bn + 1) |
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84 | 93 | buckets *= 2; |
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85 | 94 | |
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86 | 95 | /* try to allocate a large hash table to avoid collisions */ |
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87 | 96 | for (scale = 4; scale; scale /= 2) { |
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88 | 97 | h = (struct pos *)malloc(scale * buckets * sizeof(struct pos)); |
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89 | 98 | if (h) |
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90 | 99 | break; |
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91 | 100 | } |
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92 | 101 | |
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93 | 102 | if (!h) |
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94 | 103 | return 0; |
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95 | 104 | |
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96 | 105 | buckets = buckets * scale - 1; |
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97 | 106 | |
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98 | 107 | /* clear the hash table */ |
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99 | 108 | for (i = 0; i <= buckets; i++) { |
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100 | 109 | h[i].pos = -1; |
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101 | 110 | h[i].len = 0; |
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102 | 111 | } |
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103 | 112 | |
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104 | 113 | /* add lines to the hash table chains */ |
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105 | 114 | for (i = 0; i < bn; i++) { |
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106 | 115 | /* find the equivalence class */ |
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107 | 116 | for (j = b[i].hash & buckets; h[j].pos != -1; |
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108 | 117 | j = (j + 1) & buckets) |
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109 | 118 | if (!cmp(b + i, b + h[j].pos)) |
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110 | 119 | break; |
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111 | 120 | |
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112 | 121 | /* add to the head of the equivalence class */ |
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113 | 122 | b[i].n = h[j].pos; |
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114 | 123 | b[i].e = j; |
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115 | 124 | h[j].pos = i; |
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116 | 125 | h[j].len++; /* keep track of popularity */ |
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117 | 126 | } |
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118 | 127 | |
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119 | 128 | /* compute popularity threshold */ |
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120 | 129 | t = (bn >= 31000) ? bn / 1000 : 1000000 / (bn + 1); |
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121 | 130 | |
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122 | 131 | /* match items in a to their equivalence class in b */ |
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123 | 132 | for (i = 0; i < an; i++) { |
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124 | 133 | /* find the equivalence class */ |
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125 | 134 | for (j = a[i].hash & buckets; h[j].pos != -1; |
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126 | 135 | j = (j + 1) & buckets) |
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127 | 136 | if (!cmp(a + i, b + h[j].pos)) |
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128 | 137 | break; |
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129 | 138 | |
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130 | 139 | a[i].e = j; /* use equivalence class for quick compare */ |
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131 | 140 | if (h[j].len <= t) |
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132 | 141 | a[i].n = h[j].pos; /* point to head of match list */ |
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133 | 142 | else |
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134 | 143 | a[i].n = -1; /* too popular */ |
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135 | 144 | } |
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136 | 145 | |
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137 | 146 | /* discard hash tables */ |
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138 | 147 | free(h); |
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139 | 148 | return 1; |
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140 | 149 | } |
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141 | 150 | |
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142 | 151 | static int longest_match(struct bdiff_line *a, struct bdiff_line *b, |
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143 | 152 | struct pos *pos, |
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144 | 153 | int a1, int a2, int b1, int b2, int *omi, int *omj) |
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145 | 154 | { |
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146 | 155 | int mi = a1, mj = b1, mk = 0, i, j, k, half, bhalf; |
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147 | 156 | |
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148 | 157 | /* window our search on large regions to better bound |
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149 | 158 | worst-case performance. by choosing a window at the end, we |
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150 | 159 | reduce skipping overhead on the b chains. */ |
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151 | 160 | if (a2 - a1 > 30000) |
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152 | 161 | a1 = a2 - 30000; |
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153 | 162 | |
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154 | 163 | half = (a1 + a2 - 1) / 2; |
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155 | 164 | bhalf = (b1 + b2 - 1) / 2; |
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156 | 165 | |
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157 | 166 | for (i = a1; i < a2; i++) { |
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158 | 167 | /* skip all lines in b after the current block */ |
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159 | 168 | for (j = a[i].n; j >= b2; j = b[j].n) |
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160 | 169 | ; |
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161 | 170 | |
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162 | 171 | /* loop through all lines match a[i] in b */ |
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163 | 172 | for (; j >= b1; j = b[j].n) { |
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164 | 173 | /* does this extend an earlier match? */ |
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165 | 174 | for (k = 1; j - k >= b1 && i - k >= a1; k++) { |
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166 | 175 | /* reached an earlier match? */ |
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167 | 176 | if (pos[j - k].pos == i - k) { |
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168 | 177 | k += pos[j - k].len; |
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169 | 178 | break; |
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170 | 179 | } |
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171 | 180 | /* previous line mismatch? */ |
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172 | 181 | if (a[i - k].e != b[j - k].e) |
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173 | 182 | break; |
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174 | 183 | } |
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175 | 184 | |
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176 | 185 | pos[j].pos = i; |
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177 | 186 | pos[j].len = k; |
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178 | 187 | |
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179 | 188 | /* best match so far? we prefer matches closer |
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180 | 189 | to the middle to balance recursion */ |
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181 | 190 | if (k > mk) { |
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182 | 191 | /* a longer match */ |
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183 | 192 | mi = i; |
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184 | 193 | mj = j; |
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185 | 194 | mk = k; |
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186 | 195 | } else if (k == mk) { |
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187 | 196 | if (i > mi && i <= half && j > b1) { |
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188 | 197 | /* same match but closer to half */ |
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189 | 198 | mi = i; |
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190 | 199 | mj = j; |
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191 | 200 | } else if (i == mi && (mj > bhalf || i == a1)) { |
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192 | 201 | /* same i but best earlier j */ |
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193 | 202 | mj = j; |
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194 | 203 | } |
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195 | 204 | } |
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196 | 205 | } |
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197 | 206 | } |
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198 | 207 | |
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199 | 208 | if (mk) { |
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200 | 209 | mi = mi - mk + 1; |
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201 | 210 | mj = mj - mk + 1; |
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202 | 211 | } |
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203 | 212 | |
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204 | 213 | /* expand match to include subsequent popular lines */ |
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205 | 214 | while (mi + mk < a2 && mj + mk < b2 && |
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206 | 215 | a[mi + mk].e == b[mj + mk].e) |
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207 | 216 | mk++; |
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208 | 217 | |
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209 | 218 | *omi = mi; |
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210 | 219 | *omj = mj; |
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211 | 220 | |
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212 | 221 | return mk; |
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213 | 222 | } |
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214 | 223 | |
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215 | 224 | static struct bdiff_hunk *recurse(struct bdiff_line *a, struct bdiff_line *b, |
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216 | 225 | struct pos *pos, |
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217 | 226 | int a1, int a2, int b1, int b2, struct bdiff_hunk *l) |
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218 | 227 | { |
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219 | 228 | int i, j, k; |
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220 | 229 | |
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221 | 230 | while (1) { |
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222 | 231 | /* find the longest match in this chunk */ |
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223 | 232 | k = longest_match(a, b, pos, a1, a2, b1, b2, &i, &j); |
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224 | 233 | if (!k) |
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225 | 234 | return l; |
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226 | 235 | |
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227 | 236 | /* and recurse on the remaining chunks on either side */ |
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228 | 237 | l = recurse(a, b, pos, a1, i, b1, j, l); |
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229 | 238 | if (!l) |
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230 | 239 | return NULL; |
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231 | 240 | |
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232 | 241 | l->next = (struct bdiff_hunk *)malloc(sizeof(struct bdiff_hunk)); |
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233 | 242 | if (!l->next) |
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234 | 243 | return NULL; |
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235 | 244 | |
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236 | 245 | l = l->next; |
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237 | 246 | l->a1 = i; |
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238 | 247 | l->a2 = i + k; |
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239 | 248 | l->b1 = j; |
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240 | 249 | l->b2 = j + k; |
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241 | 250 | l->next = NULL; |
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242 | 251 | |
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243 | 252 | /* tail-recursion didn't happen, so do equivalent iteration */ |
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244 | 253 | a1 = i + k; |
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245 | 254 | b1 = j + k; |
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246 | 255 | } |
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247 | 256 | } |
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248 | 257 | |
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249 | 258 | int bdiff_diff(struct bdiff_line *a, int an, struct bdiff_line *b, |
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250 | 259 | int bn, struct bdiff_hunk *base) |
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251 | 260 | { |
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252 | 261 | struct bdiff_hunk *curr; |
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253 | 262 | struct pos *pos; |
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254 | 263 | int t, count = 0; |
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255 | 264 | |
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256 | 265 | /* allocate and fill arrays */ |
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257 | 266 | t = equatelines(a, an, b, bn); |
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258 | 267 | pos = (struct pos *)calloc(bn ? bn : 1, sizeof(struct pos)); |
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259 | 268 | |
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260 | 269 | if (pos && t) { |
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261 | 270 | /* generate the matching block list */ |
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262 | 271 | |
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263 | 272 | curr = recurse(a, b, pos, 0, an, 0, bn, base); |
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264 | 273 | if (!curr) |
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265 | 274 | return -1; |
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266 | 275 | |
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267 | 276 | /* sentinel end hunk */ |
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268 | 277 | curr->next = (struct bdiff_hunk *)malloc(sizeof(struct bdiff_hunk)); |
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269 | 278 | if (!curr->next) |
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270 | 279 | return -1; |
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271 | 280 | curr = curr->next; |
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272 | 281 | curr->a1 = curr->a2 = an; |
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273 | 282 | curr->b1 = curr->b2 = bn; |
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274 | 283 | curr->next = NULL; |
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275 | 284 | } |
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276 | 285 | |
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277 | 286 | free(pos); |
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278 | 287 | |
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279 | 288 | /* normalize the hunk list, try to push each hunk towards the end */ |
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280 | 289 | for (curr = base->next; curr; curr = curr->next) { |
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281 | 290 | struct bdiff_hunk *next = curr->next; |
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282 | 291 | |
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283 | 292 | if (!next) |
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284 | 293 | break; |
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285 | 294 | |
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286 | 295 | if (curr->a2 == next->a1 || curr->b2 == next->b1) |
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287 | 296 | while (curr->a2 < an && curr->b2 < bn |
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288 | 297 | && next->a1 < next->a2 |
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289 | 298 | && next->b1 < next->b2 |
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290 | 299 | && !cmp(a + curr->a2, b + curr->b2)) { |
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291 | 300 | curr->a2++; |
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292 | 301 | next->a1++; |
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293 | 302 | curr->b2++; |
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294 | 303 | next->b1++; |
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295 | 304 | } |
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296 | 305 | } |
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297 | 306 | |
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298 | 307 | for (curr = base->next; curr; curr = curr->next) |
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299 | 308 | count++; |
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300 | 309 | return count; |
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301 | 310 | } |
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302 | 311 | |
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303 | 312 | void bdiff_freehunks(struct bdiff_hunk *l) |
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304 | 313 | { |
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305 | 314 | struct bdiff_hunk *n; |
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306 | 315 | for (; l; l = n) { |
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307 | 316 | n = l->next; |
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308 | 317 | free(l); |
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309 | 318 | } |
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310 | 319 | } |
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311 | 320 | |
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312 | 321 |
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