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| 1 | 1 |
| 2 /* | 2 /* |
| 3 * Copyright 2012 Google Inc. | 3 * Copyright 2012 Google Inc. |
| 4 * | 4 * |
| 5 * Use of this source code is governed by a BSD-style license that can be | 5 * Use of this source code is governed by a BSD-style license that can be |
| 6 * found in the LICENSE file. | 6 * found in the LICENSE file. |
| 7 */ | 7 */ |
| 8 | 8 |
| 9 #include "SkTileGrid.h" | 9 #include "SkTileGrid.h" |
| 10 | 10 |
| (...skipping 48 matching lines...) Expand 10 before | Expand all | Expand 10 after Loading... | |
| 59 this->tile(x, y).push(data); | 59 this->tile(x, y).push(data); |
| 60 } | 60 } |
| 61 } | 61 } |
| 62 fInsertionCount++; | 62 fInsertionCount++; |
| 63 } | 63 } |
| 64 | 64 |
| 65 void SkTileGrid::search(const SkIRect& query, SkTDArray<void*>* results) { | 65 void SkTileGrid::search(const SkIRect& query, SkTDArray<void*>* results) { |
| 66 SkIRect adjustedQuery = query; | 66 SkIRect adjustedQuery = query; |
| 67 adjustedQuery.inset(fInfo.fMargin.width(), fInfo.fMargin.height()); | 67 adjustedQuery.inset(fInfo.fMargin.width(), fInfo.fMargin.height()); |
| 68 adjustedQuery.offset(fInfo.fOffset); | 68 adjustedQuery.offset(fInfo.fOffset); |
| 69 adjustedQuery.sort(); // in case the inset inverted the rectangle | |
|
reed1
2013/04/09 15:56:16
If we might invert, then (I presume) we are making
| |
| 69 // Convert the query rectangle from device coordinates to tile coordinates | 70 // Convert the query rectangle from device coordinates to tile coordinates |
| 70 // by rounding outwards to the nearest tile boundary so that the resulting t ile | 71 // by rounding outwards to the nearest tile boundary so that the resulting t ile |
| 71 // region includes the query rectangle. (using truncating division to "floor ") | 72 // region includes the query rectangle. (using truncating division to "floor ") |
| 72 int tileStartX = adjustedQuery.left() / fInfo.fTileInterval.width(); | 73 int tileStartX = adjustedQuery.left() / fInfo.fTileInterval.width(); |
| 73 int tileEndX = (adjustedQuery.right() + fInfo.fTileInterval.width() - 1) / | 74 int tileEndX = (adjustedQuery.right() + fInfo.fTileInterval.width() - 1) / |
| 74 fInfo.fTileInterval.width(); | 75 fInfo.fTileInterval.width(); |
| 75 int tileStartY = adjustedQuery.top() / fInfo.fTileInterval.height(); | 76 int tileStartY = adjustedQuery.top() / fInfo.fTileInterval.height(); |
| 76 int tileEndY = (adjustedQuery.bottom() + fInfo.fTileInterval.height() - 1) / | 77 int tileEndY = (adjustedQuery.bottom() + fInfo.fTileInterval.height() - 1) / |
| 77 fInfo.fTileInterval.height(); | 78 fInfo.fTileInterval.height(); |
| 78 if (tileStartX >= fXTileCount || tileStartY >= fYTileCount || tileEndX <= 0 || tileEndY <= 0) { | 79 |
| 79 return; // query does not intersect the grid | 80 tileStartX = SkPin32(tileStartX, 0, fXTileCount - 1); |
| 80 } | 81 tileEndX = SkPin32(tileEndX, 1, fXTileCount); |
| 81 // clamp to grid | 82 tileStartY = SkPin32(tileStartY, 0, fYTileCount - 1); |
| 82 if (tileStartX < 0) tileStartX = 0; | 83 tileEndY = SkPin32(tileEndY, 1, fYTileCount); |
| 83 if (tileStartY < 0) tileStartY = 0; | |
| 84 if (tileEndX > fXTileCount) tileEndX = fXTileCount; | |
| 85 if (tileEndY > fYTileCount) tileEndY = fYTileCount; | |
| 86 | 84 |
| 87 int queryTileCount = (tileEndX - tileStartX) * (tileEndY - tileStartY); | 85 int queryTileCount = (tileEndX - tileStartX) * (tileEndY - tileStartY); |
| 86 SkASSERT(queryTileCount); | |
| 88 if (queryTileCount == 1) { | 87 if (queryTileCount == 1) { |
| 89 *results = this->tile(tileStartX, tileStartY); | 88 *results = this->tile(tileStartX, tileStartY); |
| 90 } else { | 89 } else { |
| 91 results->reset(); | 90 results->reset(); |
| 92 SkTDArray<int> curPositions; | 91 SkTDArray<int> curPositions; |
| 93 curPositions.setCount(queryTileCount); | 92 curPositions.setCount(queryTileCount); |
| 94 // Note: Reserving space for 1024 tile pointers on the stack. If the | 93 // Note: Reserving space for 1024 tile pointers on the stack. If the |
| 95 // malloc becomes a bottleneck, we may consider increasing that number. | 94 // malloc becomes a bottleneck, we may consider increasing that number. |
| 96 // Typical large web page, say 2k x 16k, would require 512 tiles of | 95 // Typical large web page, say 2k x 16k, would require 512 tiles of |
| 97 // size 256 x 256 pixels. | 96 // size 256 x 256 pixels. |
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| 123 } | 122 } |
| 124 | 123 |
| 125 void SkTileGrid::rewindInserts() { | 124 void SkTileGrid::rewindInserts() { |
| 126 SkASSERT(fClient); | 125 SkASSERT(fClient); |
| 127 for (int i = 0; i < fTileCount; ++i) { | 126 for (int i = 0; i < fTileCount; ++i) { |
| 128 while (!fTileData[i].isEmpty() && fClient->shouldRewind(fTileData[i].top ())) { | 127 while (!fTileData[i].isEmpty() && fClient->shouldRewind(fTileData[i].top ())) { |
| 129 fTileData[i].pop(); | 128 fTileData[i].pop(); |
| 130 } | 129 } |
| 131 } | 130 } |
| 132 } | 131 } |
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