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@@ -0,0 +1,29 | |||||
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1 | #require no-pure | |||
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2 | ||||
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3 | A script to generate nasty diff worst-case scenarios: | |||
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4 | ||||
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5 | $ cat > s.py <<EOF | |||
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6 | > import random | |||
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7 | > for x in xrange(100000): | |||
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8 | ||||
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9 | > if random.randint(0, 100) >= 50: | |||
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10 | > x += 1 | |||
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11 | > print hex(x) | |||
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12 | > EOF | |||
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13 | ||||
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14 | $ hg init a | |||
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15 | $ cd a | |||
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16 | ||||
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17 | Check in a big file: | |||
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18 | ||||
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19 | $ python ../s.py > a | |||
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20 | $ hg ci -qAm0 | |||
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21 | ||||
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22 | Modify it: | |||
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23 | ||||
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24 | $ python ../s.py > a | |||
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25 | ||||
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26 | Time a check-in, should never take more than 10 seconds user time: | |||
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27 | ||||
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28 | $ hg ci --time -m1 | |||
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29 | time: real .* secs .user [0-9][.].* sys .* (re) |
@@ -148,7 +148,7 static int equatelines(struct line *a, i | |||||
148 | static int longest_match(struct line *a, struct line *b, struct pos *pos, |
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148 | static int longest_match(struct line *a, struct line *b, struct pos *pos, | |
149 | int a1, int a2, int b1, int b2, int *omi, int *omj) |
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149 | int a1, int a2, int b1, int b2, int *omi, int *omj) | |
150 | { |
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150 | { | |
151 | int mi = a1, mj = b1, mk = 0, mb = 0, i, j, k; |
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151 | int mi = a1, mj = b1, mk = 0, mb = 0, i, j, k, half = (a1 + a2) / 2; | |
152 |
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152 | |||
153 | for (i = a1; i < a2; i++) { |
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153 | for (i = a1; i < a2; i++) { | |
154 | /* skip all lines in b after the current block */ |
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154 | /* skip all lines in b after the current block */ | |
@@ -165,8 +165,9 static int longest_match(struct line *a, | |||||
165 | pos[j].pos = i; |
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165 | pos[j].pos = i; | |
166 | pos[j].len = k; |
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166 | pos[j].len = k; | |
167 |
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167 | |||
168 |
/* best match so far? |
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168 | /* best match so far? we prefer matches closer | |
169 | if (k > mk || (k == mk && i <= mi)) { |
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169 | to the middle to balance recursion */ | |
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170 | if (k > mk || (k == mk && (i <= mi || i < half))) { | |||
170 | mi = i; |
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171 | mi = i; | |
171 | mj = j; |
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172 | mj = j; | |
172 | mk = k; |
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173 | mk = k; |
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