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 HistogramDiff} is used as diff algorithm.
61 */
62 public final class MergeAlgorithm {
63 private final DiffAlgorithm diffAlg;
64
65 /**
66 * Creates a new MergeAlgorithm which uses {@link HistogramDiff} as diff
67 * algorithm
68 */
69 public MergeAlgorithm() {
70 this(new HistogramDiff());
71 }
72
73 /**
74 * Creates a new MergeAlgorithm
75 *
76 * @param diff
77 * the diff algorithm used by this merge
78 */
79 public MergeAlgorithm(DiffAlgorithm diff) {
80 this.diffAlg = diff;
81 }
82
83 // An special edit which acts as a sentinel value by marking the end the
84 // list of edits
85 private final static Edit END_EDIT = new Edit(Integer.MAX_VALUE,
86 Integer.MAX_VALUE);
87
88 /**
89 * Does the three way merge between a common base and two sequences.
90 *
91 * @param <S>
92 * type of sequence.
93 * @param cmp comparison method for this execution.
94 * @param base the common base sequence
95 * @param ours the first sequence to be merged
96 * @param theirs the second sequence to be merged
97 * @return the resulting content
98 */
99 public <S extends Sequence> MergeResult<S> merge(
100 SequenceComparator<S> cmp, S base, S ours, S theirs) {
101 List<S> sequences = new ArrayList<S>(3);
102 sequences.add(base);
103 sequences.add(ours);
104 sequences.add(theirs);
105 MergeResult<S> result = new MergeResult<S>(sequences);
106
107 if (ours.size() == 0) {
108 if (theirs.size() != 0) {
109 EditList theirsEdits = diffAlg.diff(cmp, base, theirs);
110 if (!theirsEdits.isEmpty()) {
111 // we deleted, they modified -> Let their complete content
112 // conflict with empty text
113 result.add(1, 0, 0, ConflictState.FIRST_CONFLICTING_RANGE);
114 result.add(2, 0, theirs.size(),
115 ConflictState.NEXT_CONFLICTING_RANGE);
116 } else
117 // we deleted, they didn't modify -> Let our deletion win
118 result.add(1, 0, 0, ConflictState.NO_CONFLICT);
119 } else
120 // we and they deleted -> return a single chunk of nothing
121 result.add(1, 0, 0, ConflictState.NO_CONFLICT);
122 return result;
123 } else if (theirs.size() == 0) {
124 EditList oursEdits = diffAlg.diff(cmp, base, ours);
125 if (!oursEdits.isEmpty()) {
126 // we modified, they deleted -> Let our complete content
127 // conflict with empty text
128 result.add(1, 0, ours.size(),
129 ConflictState.FIRST_CONFLICTING_RANGE);
130 result.add(2, 0, 0, ConflictState.NEXT_CONFLICTING_RANGE);
131 } else
132 // they deleted, we didn't modify -> Let their deletion win
133 result.add(2, 0, 0, ConflictState.NO_CONFLICT);
134 return result;
135 }
136
137 EditList oursEdits = diffAlg.diff(cmp, base, ours);
138 Iterator<Edit> baseToOurs = oursEdits.iterator();
139 EditList theirsEdits = diffAlg.diff(cmp, base, theirs);
140 Iterator<Edit> baseToTheirs = theirsEdits.iterator();
141 int current = 0; // points to the next line (first line is 0) of base
142 // which was not handled yet
143 Edit oursEdit = nextEdit(baseToOurs);
144 Edit theirsEdit = nextEdit(baseToTheirs);
145
146 // iterate over all edits from base to ours and from base to theirs
147 // leave the loop when there are no edits more for ours or for theirs
148 // (or both)
149 while (theirsEdit != END_EDIT || oursEdit != END_EDIT) {
150 if (oursEdit.getEndA() < theirsEdit.getBeginA()) {
151 // something was changed in ours not overlapping with any change
152 // from theirs. First add the common part in front of the edit
153 // then the edit.
154 if (current != oursEdit.getBeginA()) {
155 result.add(0, current, oursEdit.getBeginA(),
156 ConflictState.NO_CONFLICT);
157 }
158 result.add(1, oursEdit.getBeginB(), oursEdit.getEndB(),
159 ConflictState.NO_CONFLICT);
160 current = oursEdit.getEndA();
161 oursEdit = nextEdit(baseToOurs);
162 } else if (theirsEdit.getEndA() < oursEdit.getBeginA()) {
163 // something was changed in theirs not overlapping with any
164 // from ours. First add the common part in front of the edit
165 // then the edit.
166 if (current != theirsEdit.getBeginA()) {
167 result.add(0, current, theirsEdit.getBeginA(),
168 ConflictState.NO_CONFLICT);
169 }
170 result.add(2, theirsEdit.getBeginB(), theirsEdit.getEndB(),
171 ConflictState.NO_CONFLICT);
172 current = theirsEdit.getEndA();
173 theirsEdit = nextEdit(baseToTheirs);
174 } else {
175 // here we found a real overlapping modification
176
177 // if there is a common part in front of the conflict add it
178 if (oursEdit.getBeginA() != current
179 && theirsEdit.getBeginA() != current) {
180 result.add(0, current, Math.min(oursEdit.getBeginA(),
181 theirsEdit.getBeginA()), ConflictState.NO_CONFLICT);
182 }
183
184 // set some initial values for the ranges in A and B which we
185 // want to handle
186 int oursBeginB = oursEdit.getBeginB();
187 int theirsBeginB = theirsEdit.getBeginB();
188 // harmonize the start of the ranges in A and B
189 if (oursEdit.getBeginA() < theirsEdit.getBeginA()) {
190 theirsBeginB -= theirsEdit.getBeginA()
191 - oursEdit.getBeginA();
192 } else {
193 oursBeginB -= oursEdit.getBeginA() - theirsEdit.getBeginA();
194 }
195
196 // combine edits:
197 // Maybe an Edit on one side corresponds to multiple Edits on
198 // the other side. Then we have to combine the Edits of the
199 // other side - so in the end we can merge together two single
200 // edits.
201 //
202 // It is important to notice that this combining will extend the
203 // ranges of our conflict always downwards (towards the end of
204 // the content). The starts of the conflicting ranges in ours
205 // and theirs are not touched here.
206 //
207 // This combining is an iterative process: after we have
208 // combined some edits we have to do the check again. The
209 // combined edits could now correspond to multiple edits on the
210 // other side.
211 //
212 // Example: when this combining algorithm works on the following
213 // edits
214 // oursEdits=((0-5,0-5),(6-8,6-8),(10-11,10-11)) and
215 // theirsEdits=((0-1,0-1),(2-3,2-3),(5-7,5-7))
216 // it will merge them into
217 // oursEdits=((0-8,0-8),(10-11,10-11)) and
218 // theirsEdits=((0-7,0-7))
219 //
220 // Since the only interesting thing to us is how in ours and
221 // theirs the end of the conflicting range is changing we let
222 // oursEdit and theirsEdit point to the last conflicting edit
223 Edit nextOursEdit = nextEdit(baseToOurs);
224 Edit nextTheirsEdit = nextEdit(baseToTheirs);
225 for (;;) {
226 if (oursEdit.getEndA() >= nextTheirsEdit.getBeginA()) {
227 theirsEdit = nextTheirsEdit;
228 nextTheirsEdit = nextEdit(baseToTheirs);
229 } else if (theirsEdit.getEndA() >= nextOursEdit.getBeginA()) {
230 oursEdit = nextOursEdit;
231 nextOursEdit = nextEdit(baseToOurs);
232 } else {
233 break;
234 }
235 }
236
237 // harmonize the end of the ranges in A and B
238 int oursEndB = oursEdit.getEndB();
239 int theirsEndB = theirsEdit.getEndB();
240 if (oursEdit.getEndA() < theirsEdit.getEndA()) {
241 oursEndB += theirsEdit.getEndA() - oursEdit.getEndA();
242 } else {
243 theirsEndB += oursEdit.getEndA() - theirsEdit.getEndA();
244 }
245
246 // A conflicting region is found. Strip off common lines in
247 // in the beginning and the end of the conflicting region
248
249 // Determine the minimum length of the conflicting areas in OURS
250 // and THEIRS. Also determine how much bigger the conflicting
251 // area in THEIRS is compared to OURS. All that is needed to
252 // limit the search for common areas at the beginning or end
253 // (the common areas cannot be bigger then the smaller
254 // conflicting area. The delta is needed to know whether the
255 // complete conflicting area is common in OURS and THEIRS.
256 int minBSize = oursEndB - oursBeginB;
257 int BSizeDelta = minBSize - (theirsEndB - theirsBeginB);
258 if (BSizeDelta > 0)
259 minBSize -= BSizeDelta;
260
261 int commonPrefix = 0;
262 while (commonPrefix < minBSize
263 && cmp.equals(ours, oursBeginB + commonPrefix, theirs,
264 theirsBeginB + commonPrefix))
265 commonPrefix++;
266 minBSize -= commonPrefix;
267 int commonSuffix = 0;
268 while (commonSuffix < minBSize
269 && cmp.equals(ours, oursEndB - commonSuffix - 1, theirs,
270 theirsEndB - commonSuffix - 1))
271 commonSuffix++;
272 minBSize -= commonSuffix;
273
274 // Add the common lines at start of conflict
275 if (commonPrefix > 0)
276 result.add(1, oursBeginB, oursBeginB + commonPrefix,
277 ConflictState.NO_CONFLICT);
278
279 // Add the conflict (Only if there is a conflict left to report)
280 if (minBSize > 0 || BSizeDelta != 0) {
281 result.add(1, oursBeginB + commonPrefix, oursEndB
282 - commonSuffix,
283 ConflictState.FIRST_CONFLICTING_RANGE);
284 result.add(2, theirsBeginB + commonPrefix, theirsEndB
285 - commonSuffix,
286 ConflictState.NEXT_CONFLICTING_RANGE);
287 }
288
289 // Add the common lines at end of conflict
290 if (commonSuffix > 0)
291 result.add(1, oursEndB - commonSuffix, oursEndB,
292 ConflictState.NO_CONFLICT);
293
294 current = Math.max(oursEdit.getEndA(), theirsEdit.getEndA());
295 oursEdit = nextOursEdit;
296 theirsEdit = nextTheirsEdit;
297 }
298 }
299 // maybe we have a common part behind the last edit: copy it to the
300 // result
301 if (current < base.size()) {
302 result.add(0, current, base.size(), ConflictState.NO_CONFLICT);
303 }
304 return result;
305 }
306
307 /**
308 * Helper method which returns the next Edit for an Iterator over Edits.
309 * When there are no more edits left this method will return the constant
310 * END_EDIT.
311 *
312 * @param it
313 * the iterator for which the next edit should be returned
314 * @return the next edit from the iterator or END_EDIT if there no more
315 * edits
316 */
317 private static Edit nextEdit(Iterator<Edit> it) {
318 return (it.hasNext() ? it.next() : END_EDIT);
319 }
320 }