Table with fixed layout behaves like auto layout when its width is set by js instead of css.

Source/core/rendering/RenderTable.cpp

Index: Source/core/rendering/RenderTable.cpp
diff --git a/Source/core/rendering/RenderTable.cpp b/Source/core/rendering/RenderTable.cpp
index b1dfd7ee6304aad949a29421810b0daa77db4f10..90f95d7b9e33a1eba02a5011611e0344fa7652fe 100644
--- a/Source/core/rendering/RenderTable.cpp
+++ b/Source/core/rendering/RenderTable.cpp
@@ -90,7 +90,7 @@ void RenderTable::styleDidChange(StyleDifference diff, const RenderStyle* oldSty
     m_vSpacing = collapseBorders() ? 0 : style()->verticalBorderSpacing();
     m_columnPos[0] = m_hSpacing;
 
-    if (!m_tableLayout || style()->tableLayout() != oldTableLayout) {
+    if (!m_tableLayout || style()->tableLayout() != oldTableLayout || diff.needsFullLayout()) {

[mstensho (USE GERRIT), 2014/05/06 13:28:59]

It seems wrong to recreate the table layout object every time you want to lay
out the table. You only want to do this if it's missing or if you're really
going to switch layout algorithm.

The problem here is the "style()->tableLayout() != oldTableLayout" test. It
doesn't account for the fact that auto width means auto table layout, even if
the specified value of 'table-layout' is 'fixed'.

A helper method (e.g. "isTableLayoutFixed()") in RenderStyle, which checks both
for table-layout:fixed AND non-auto width, could be a better solution. It could
be called both on this line, and on a few lines further below (which already
does the right thing).

         if (m_tableLayout)
             m_tableLayout->willChangeTableLayout();