Tweak out SkTileGrid::insert() loop.

src/core/SkTileGrid.cpp

 /*
  * Copyright 2012 Google Inc.
  *
  * Use of this source code is governed by a BSD-style license that can be
  * found in the LICENSE file.
  */
 
 #include "SkTileGrid.h"
 
 SkTileGrid::SkTileGrid(int xTiles, int yTiles, const SkTileGridFactory::TileGridInfo& info)
@@ skipping 66 matching lines @@
 
     // TODO(mtklein): can we assert this instead to save an intersection in Release mode,
     // or just allow out-of-bound insertions to insert anyway (clamped to nearest tile)?
     if (!SkRect::Intersects(bounds, fGridBounds)) {
         return;
     }
 
     SkIRect grid;
     this->userToGrid(bounds, &grid);
 
-    for (int y = grid.fTop; y <= grid.fBottom; y++) {
-        for (int x = grid.fLeft; x <= grid.fRight; x++) {
-            fTiles[y * fXTiles + x].push(opIndex);
+    // This is just a loop over y then x.  This compiles to a slightly faster and
+    // more compact loop than if we just did fTiles[y * fXTiles + x].push(opIndex).
+    SkTDArray<unsigned>* row = &fTiles[grid.fTop * fXTiles + grid.fLeft];
+    for (int y = 0; y <= grid.fBottom - grid.fTop; y++) {
+        SkTDArray<unsigned>* tile = row;
+        for (int x = 0; x <= grid.fRight - grid.fLeft; x++) {
+            (tile++)->push(opIndex);
         }
+        row += fXTiles;
     }
 }
 
 // Number of tiles for which data is allocated on the stack in
 // SkTileGrid::search. If malloc becomes a bottleneck, we may consider
 // increasing this number. Typical large web page, say 2k x 16k, would
 // require 512 tiles of size 256 x 256 pixels.
 static const int kStackAllocationTileCount = 1024;
 
 void SkTileGrid::search(const SkRect& originalQuery, SkTDArray<unsigned>* results) const {
@@ skipping 57 matching lines @@
         // We did find an earliest op. Output it, and step forward every tile that contains it.
         results->push(earliest);
         for (int i = 0; i < starts.count(); i++) {
             if (starts[i] < ends[i] && *starts[i] == earliest) {
                 starts[i]++;
             }
         }
     }
 }