Remove clips where we can show that they are unnecessary.

Source/core/platform/graphics/RoundedRect.cpp

Index: Source/core/platform/graphics/RoundedRect.cpp
diff --git a/Source/core/platform/graphics/RoundedRect.cpp b/Source/core/platform/graphics/RoundedRect.cpp
index 124af84a62ef7880bd0681673d5d9d0f186eea01..804eb0ad0d081c893da64baa1c490400b43ada53 100644
--- a/Source/core/platform/graphics/RoundedRect.cpp
+++ b/Source/core/platform/graphics/RoundedRect.cpp
@@ -80,6 +80,15 @@ void RoundedRect::Radii::expand(int topWidth, int bottomWidth, int leftWidth, in
     }
 }
 
+IntRect RoundedRect::enclosedRect() const

[Julien - ping for review, 2013/06/19 19:42:58]

This will return a smaller rectangle that what enclosed rect would be expected
to return AFAICT. If you look at
http://www.w3.org/TR/css3-background/#border-radius, you are returning the rect
formed by all the ellipses centers.

[jbroman, 2013/08/02 14:28:17]

What I want is "quickly give me a rect which is enclosed by this rounded rect"
(the idea being to be able to say that if some geometry is enclosed by
rrect.enclosedRect(), then it is enclosed by rrect).

Using the ellipse centres seemed like a reasonable way of computing such a rect.
In the common case, the radii are all a few pixels and the returned rect is just
inside the outer rect (i.e. the one returned by RoundedRect::rect).

What do you suggest? A name change? A comment?

+{
+    int minX = m_rect.x() + max(m_radii.topLeft().width(), m_radii.bottomLeft().width());
+    int minY = m_rect.y() + max(m_radii.topLeft().height(), m_radii.topRight().height());
+    int maxX = m_rect.maxX() - max(m_radii.topRight().width(), m_radii.bottomRight().width());
+    int maxY = m_rect.maxY() - max(m_radii.bottomLeft().height(), m_radii.bottomRight().height());
+    return IntRect(minX, minY, maxX - minX, maxY - minY);
+}
+
 void RoundedRect::inflateWithRadii(int size)
 {
     IntRect old = m_rect;