Optimize QuadF's IsCounterClockwise function

ui/gfx/geometry/quad_f.cc

 // Copyright (c) 2012 The Chromium Authors. All rights reserved.
 // Use of this source code is governed by a BSD-style license that can be
 // found in the LICENSE file.
 
 #include "ui/gfx/geometry/quad_f.h"
 
 #include <limits>
 
 #include "base/strings/stringprintf.h"
 
@@ skipping 27 matching lines @@
 }
 
 bool QuadF::IsCounterClockwise() const {
   // This math computes the signed area of the quad. Positive area
   // indicates the quad is clockwise; negative area indicates the quad is
   // counter-clockwise. Note carefully: this is backwards from conventional
   // math because our geometric space uses screen coordiantes with y-axis
   // pointing downards.
   // Reference: http://mathworld.wolfram.com/PolygonArea.html
 
+  float tmp1 = p2_.y() - p4_.y();

[danakj, 2014/08/11 14:18:49]

how about naming this p24 or something more descriptive than "tmp"?

[kui.zheng, 2014/08/12 04:20:55]

Done.

+  float tmp2 = p3_.y() - p1_.y();

[danakj, 2014/08/11 14:18:49]

ditto

[kui.zheng, 2014/08/12 04:20:55]

Done.

+
   // Up-cast to double so this cannot overflow.
-  double determinant1 = static_cast<double>(p1_.x()) * p2_.y()
-      - static_cast<double>(p2_.x()) * p1_.y();
-  double determinant2 = static_cast<double>(p2_.x()) * p3_.y()
-      - static_cast<double>(p3_.x()) * p2_.y();
-  double determinant3 = static_cast<double>(p3_.x()) * p4_.y()
-      - static_cast<double>(p4_.x()) * p3_.y();
-  double determinant4 = static_cast<double>(p4_.x()) * p1_.y()
-      - static_cast<double>(p1_.x()) * p4_.y();
+  double determinant1 = static_cast<double>(p1_.x()) * tmp1;

[Ian Vollick, 2014/08/11 14:39:09]

If I'm understanding things, determinant1 here is no longer a determinant, but
rather computes some of the terms involved in the old code's sum. If that's
true, I'd vote for something like what Dana suggested above for tmp1 and tmp2
rather than "determinant" and maybe add a comment explaining that what's here is
equivalent to computing the sum of the determinants (i.e., that if you flip the
sign of determinant3 and determinant4 and add them to determinant1 and
determinant2 here, you get the same value as before).

[kui.zheng, 2014/08/12 04:20:55]

Right, This patch just refactoring the computation of adding determinants so
that reducing the number of operations. So, "determinant" is meaningless here. 

+  double determinant2 = static_cast<double>(p2_.x()) * tmp2;
+  double determinant3 = static_cast<double>(p3_.x()) * tmp1;
+  double determinant4 = static_cast<double>(p4_.x()) * tmp2;
 
-  return determinant1 + determinant2 + determinant3 + determinant4 < 0;
+  return determinant1 + determinant2 < determinant3 + determinant4;
 }
 
 static inline bool PointIsInTriangle(const PointF& point,
                                      const PointF& r1,
                                      const PointF& r2,
                                      const PointF& r3) {
   // Compute the barycentric coordinates (u, v, w) of |point| relative to the
   // triangle (r1, r2, r3) by the solving the system of equations:
   //   1) point = u * r1 + v * r2 + w * r3
   //   2) u + v + w = 1
@@ skipping 45 matching lines @@
   return result;
 }
 
 QuadF operator-(const QuadF& lhs, const Vector2dF& rhs) {
   QuadF result = lhs;
   result -= rhs;
   return result;
 }
 
 }  // namespace gfx