1 /*
2 * Copyright (C) 2009, Christian Halstrick <christian.halstrick@sap.com>
3 * and other copyright owners as documented in the project's IP log.
4 *
5 * This program and the accompanying materials are made available
6 * under the terms of the Eclipse Distribution License v1.0 which
7 * accompanies this distribution, is reproduced below, and is
8 * available at http://www.eclipse.org/org/documents/edl-v10.php
9 *
10 * All rights reserved.
11 *
12 * Redistribution and use in source and binary forms, with or
13 * without modification, are permitted provided that the following
14 * conditions are met:
15 *
16 * - Redistributions of source code must retain the above copyright
17 * notice, this list of conditions and the following disclaimer.
18 *
19 * - Redistributions in binary form must reproduce the above
20 * copyright notice, this list of conditions and the following
21 * disclaimer in the documentation and/or other materials provided
22 * with the distribution.
23 *
24 * - Neither the name of the Eclipse Foundation, Inc. nor the
25 * names of its contributors may be used to endorse or promote
26 * products derived from this software without specific prior
27 * written permission.
28 *
29 * THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND
30 * CONTRIBUTORS "AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES,
31 * INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES
32 * OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
33 * ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT OWNER OR
34 * CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL,
35 * SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT
36 * NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES;
37 * LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER
38 * CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT,
39 * STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
40 * ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF
41 * ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
42 */
43
44 package org.eclipse.jgit.merge;
45
46 import java.util.ArrayList;
47 import java.util.Iterator;
48 import java.util.List;
49
50 import org.eclipse.jgit.diff.DiffAlgorithm;
51 import org.eclipse.jgit.diff.Edit;
52 import org.eclipse.jgit.diff.EditList;
53 import org.eclipse.jgit.diff.HistogramDiff;
54 import org.eclipse.jgit.diff.Sequence;
55 import org.eclipse.jgit.diff.SequenceComparator;
56 import org.eclipse.jgit.merge.MergeChunk.ConflictState;
57
58 /**
59 * Provides the merge algorithm which does a three-way merge on content provided
60 * as RawText. By default {@link org.eclipse.jgit.diff.HistogramDiff} is used as
61 * diff algorithm.
62 */
63 public final class MergeAlgorithm {
64 private final DiffAlgorithm diffAlg;
65
66 /**
67 * Creates a new MergeAlgorithm which uses
68 * {@link org.eclipse.jgit.diff.HistogramDiff} as diff algorithm
69 */
70 public MergeAlgorithm() {
71 this(new HistogramDiff());
72 }
73
74 /**
75 * Creates a new MergeAlgorithm
76 *
77 * @param diff
78 * the diff algorithm used by this merge
79 */
80 public MergeAlgorithm(DiffAlgorithm diff) {
81 this.diffAlg = diff;
82 }
83
84 // An special edit which acts as a sentinel value by marking the end the
85 // list of edits
86 private final static Edit END_EDIT = new Edit(Integer.MAX_VALUE,
87 Integer.MAX_VALUE);
88
89 /**
90 * Does the three way merge between a common base and two sequences.
91 *
92 * @param cmp comparison method for this execution.
93 * @param base the common base sequence
94 * @param ours the first sequence to be merged
95 * @param theirs the second sequence to be merged
96 * @return the resulting content
97 */
98 public <S extends Sequence> MergeResult<S> merge(
99 SequenceComparator<S> cmp, S base, S ours, S theirs) {
100 List<S> sequences = new ArrayList<>(3);
101 sequences.add(base);
102 sequences.add(ours);
103 sequences.add(theirs);
104 MergeResult<S> result = new MergeResult<>(sequences);
105
106 if (ours.size() == 0) {
107 if (theirs.size() != 0) {
108 EditList theirsEdits = diffAlg.diff(cmp, base, theirs);
109 if (!theirsEdits.isEmpty()) {
110 // we deleted, they modified -> Let their complete content
111 // conflict with empty text
112 result.add(1, 0, 0, ConflictState.FIRST_CONFLICTING_RANGE);
113 result.add(2, 0, theirs.size(),
114 ConflictState.NEXT_CONFLICTING_RANGE);
115 } else
116 // we deleted, they didn't modify -> Let our deletion win
117 result.add(1, 0, 0, ConflictState.NO_CONFLICT);
118 } else
119 // we and they deleted -> return a single chunk of nothing
120 result.add(1, 0, 0, ConflictState.NO_CONFLICT);
121 return result;
122 } else if (theirs.size() == 0) {
123 EditList oursEdits = diffAlg.diff(cmp, base, ours);
124 if (!oursEdits.isEmpty()) {
125 // we modified, they deleted -> Let our complete content
126 // conflict with empty text
127 result.add(1, 0, ours.size(),
128 ConflictState.FIRST_CONFLICTING_RANGE);
129 result.add(2, 0, 0, ConflictState.NEXT_CONFLICTING_RANGE);
130 } else
131 // they deleted, we didn't modify -> Let their deletion win
132 result.add(2, 0, 0, ConflictState.NO_CONFLICT);
133 return result;
134 }
135
136 EditList oursEdits = diffAlg.diff(cmp, base, ours);
137 Iterator<Edit> baseToOurs = oursEdits.iterator();
138 EditList theirsEdits = diffAlg.diff(cmp, base, theirs);
139 Iterator<Edit> baseToTheirs = theirsEdits.iterator();
140 int current = 0; // points to the next line (first line is 0) of base
141 // which was not handled yet
142 Edit oursEdit = nextEdit(baseToOurs);
143 Edit theirsEdit = nextEdit(baseToTheirs);
144
145 // iterate over all edits from base to ours and from base to theirs
146 // leave the loop when there are no edits more for ours or for theirs
147 // (or both)
148 while (theirsEdit != END_EDIT || oursEdit != END_EDIT) {
149 if (oursEdit.getEndA() < theirsEdit.getBeginA()) {
150 // something was changed in ours not overlapping with any change
151 // from theirs. First add the common part in front of the edit
152 // then the edit.
153 if (current != oursEdit.getBeginA()) {
154 result.add(0, current, oursEdit.getBeginA(),
155 ConflictState.NO_CONFLICT);
156 }
157 result.add(1, oursEdit.getBeginB(), oursEdit.getEndB(),
158 ConflictState.NO_CONFLICT);
159 current = oursEdit.getEndA();
160 oursEdit = nextEdit(baseToOurs);
161 } else if (theirsEdit.getEndA() < oursEdit.getBeginA()) {
162 // something was changed in theirs not overlapping with any
163 // from ours. First add the common part in front of the edit
164 // then the edit.
165 if (current != theirsEdit.getBeginA()) {
166 result.add(0, current, theirsEdit.getBeginA(),
167 ConflictState.NO_CONFLICT);
168 }
169 result.add(2, theirsEdit.getBeginB(), theirsEdit.getEndB(),
170 ConflictState.NO_CONFLICT);
171 current = theirsEdit.getEndA();
172 theirsEdit = nextEdit(baseToTheirs);
173 } else {
174 // here we found a real overlapping modification
175
176 // if there is a common part in front of the conflict add it
177 if (oursEdit.getBeginA() != current
178 && theirsEdit.getBeginA() != current) {
179 result.add(0, current, Math.min(oursEdit.getBeginA(),
180 theirsEdit.getBeginA()), ConflictState.NO_CONFLICT);
181 }
182
183 // set some initial values for the ranges in A and B which we
184 // want to handle
185 int oursBeginB = oursEdit.getBeginB();
186 int theirsBeginB = theirsEdit.getBeginB();
187 // harmonize the start of the ranges in A and B
188 if (oursEdit.getBeginA() < theirsEdit.getBeginA()) {
189 theirsBeginB -= theirsEdit.getBeginA()
190 - oursEdit.getBeginA();
191 } else {
192 oursBeginB -= oursEdit.getBeginA() - theirsEdit.getBeginA();
193 }
194
195 // combine edits:
196 // Maybe an Edit on one side corresponds to multiple Edits on
197 // the other side. Then we have to combine the Edits of the
198 // other side - so in the end we can merge together two single
199 // edits.
200 //
201 // It is important to notice that this combining will extend the
202 // ranges of our conflict always downwards (towards the end of
203 // the content). The starts of the conflicting ranges in ours
204 // and theirs are not touched here.
205 //
206 // This combining is an iterative process: after we have
207 // combined some edits we have to do the check again. The
208 // combined edits could now correspond to multiple edits on the
209 // other side.
210 //
211 // Example: when this combining algorithm works on the following
212 // edits
213 // oursEdits=((0-5,0-5),(6-8,6-8),(10-11,10-11)) and
214 // theirsEdits=((0-1,0-1),(2-3,2-3),(5-7,5-7))
215 // it will merge them into
216 // oursEdits=((0-8,0-8),(10-11,10-11)) and
217 // theirsEdits=((0-7,0-7))
218 //
219 // Since the only interesting thing to us is how in ours and
220 // theirs the end of the conflicting range is changing we let
221 // oursEdit and theirsEdit point to the last conflicting edit
222 Edit nextOursEdit = nextEdit(baseToOurs);
223 Edit nextTheirsEdit = nextEdit(baseToTheirs);
224 for (;;) {
225 if (oursEdit.getEndA() >= nextTheirsEdit.getBeginA()) {
226 theirsEdit = nextTheirsEdit;
227 nextTheirsEdit = nextEdit(baseToTheirs);
228 } else if (theirsEdit.getEndA() >= nextOursEdit.getBeginA()) {
229 oursEdit = nextOursEdit;
230 nextOursEdit = nextEdit(baseToOurs);
231 } else {
232 break;
233 }
234 }
235
236 // harmonize the end of the ranges in A and B
237 int oursEndB = oursEdit.getEndB();
238 int theirsEndB = theirsEdit.getEndB();
239 if (oursEdit.getEndA() < theirsEdit.getEndA()) {
240 oursEndB += theirsEdit.getEndA() - oursEdit.getEndA();
241 } else {
242 theirsEndB += oursEdit.getEndA() - theirsEdit.getEndA();
243 }
244
245 // A conflicting region is found. Strip off common lines in
246 // in the beginning and the end of the conflicting region
247
248 // Determine the minimum length of the conflicting areas in OURS
249 // and THEIRS. Also determine how much bigger the conflicting
250 // area in THEIRS is compared to OURS. All that is needed to
251 // limit the search for common areas at the beginning or end
252 // (the common areas cannot be bigger then the smaller
253 // conflicting area. The delta is needed to know whether the
254 // complete conflicting area is common in OURS and THEIRS.
255 int minBSize = oursEndB - oursBeginB;
256 int BSizeDelta = minBSize - (theirsEndB - theirsBeginB);
257 if (BSizeDelta > 0)
258 minBSize -= BSizeDelta;
259
260 int commonPrefix = 0;
261 while (commonPrefix < minBSize
262 && cmp.equals(ours, oursBeginB + commonPrefix, theirs,
263 theirsBeginB + commonPrefix))
264 commonPrefix++;
265 minBSize -= commonPrefix;
266 int commonSuffix = 0;
267 while (commonSuffix < minBSize
268 && cmp.equals(ours, oursEndB - commonSuffix - 1, theirs,
269 theirsEndB - commonSuffix - 1))
270 commonSuffix++;
271 minBSize -= commonSuffix;
272
273 // Add the common lines at start of conflict
274 if (commonPrefix > 0)
275 result.add(1, oursBeginB, oursBeginB + commonPrefix,
276 ConflictState.NO_CONFLICT);
277
278 // Add the conflict (Only if there is a conflict left to report)
279 if (minBSize > 0 || BSizeDelta != 0) {
280 result.add(1, oursBeginB + commonPrefix, oursEndB
281 - commonSuffix,
282 ConflictState.FIRST_CONFLICTING_RANGE);
283 result.add(2, theirsBeginB + commonPrefix, theirsEndB
284 - commonSuffix,
285 ConflictState.NEXT_CONFLICTING_RANGE);
286 }
287
288 // Add the common lines at end of conflict
289 if (commonSuffix > 0)
290 result.add(1, oursEndB - commonSuffix, oursEndB,
291 ConflictState.NO_CONFLICT);
292
293 current = Math.max(oursEdit.getEndA(), theirsEdit.getEndA());
294 oursEdit = nextOursEdit;
295 theirsEdit = nextTheirsEdit;
296 }
297 }
298 // maybe we have a common part behind the last edit: copy it to the
299 // result
300 if (current < base.size()) {
301 result.add(0, current, base.size(), ConflictState.NO_CONFLICT);
302 }
303 return result;
304 }
305
306 /**
307 * Helper method which returns the next Edit for an Iterator over Edits.
308 * When there are no more edits left this method will return the constant
309 * END_EDIT.
310 *
311 * @param it
312 * the iterator for which the next edit should be returned
313 * @return the next edit from the iterator or END_EDIT if there no more
314 * edits
315 */
316 private static Edit nextEdit(Iterator<Edit> it) {
317 return (it.hasNext() ? it.next() : END_EDIT);
318 }
319 }