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			482 lines
		
	
	
		
			16 KiB
		
	
	
	
		
			HTML
		
	
	
	
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<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
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<head>
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  <title>The GNU Implementation of java.awt.geom.FlatteningPathIterator</title>
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  <meta name="author" content="Sascha Brawer" />
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  <style type="text/css"><!--
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</head>
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<body>
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<h1>The GNU Implementation of FlatteningPathIterator</h1>
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<p><i><a href="http://www.dandelis.ch/people/brawer/">Sascha
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Brawer</a>, November 2003</i></p>
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<p>This document describes the GNU implementation of the class
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<code>java.awt.geom.FlatteningPathIterator</code>. It does
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<em>not</em> describe how a programmer should use this class; please
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refer to the generated API documentation for this purpose. Instead, it
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is intended for maintenance programmers who want to understand the
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implementation, for example because they want to extend the class or
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fix a bug.</p>
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<h2>Data Structures</h2>
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<p>The algorithm uses a stack. Its allocation is delayed to the time
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when the source path iterator actually returns the first curved
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segment (either <code>SEG_QUADTO</code> or <code>SEG_CUBICTO</code>).
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If the input path does not contain any curved segments, the value of
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the <code>stack</code> variable stays <code>null</code>. In this quite
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common case, the memory consumption is minimal.</p>
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<dl><dt><code>stack</code></dt><dd>The variable <code>stack</code> is
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a <code>double</code> array that holds the start, control and end
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points of individual sub-segments.</dd>
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<dt><code>recLevel</code></dt><dd>The variable <code>recLevel</code>
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holds how many recursive sub-divisions were needed to calculate a
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segment. The original curve has recursion level 0. For each
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sub-division, the corresponding recursion level is increased by
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one.</dd>
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<dt><code>stackSize</code></dt><dd>Finally, the variable
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<code>stackSize</code> indicates how many sub-segments are stored on
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the stack.</dd></dl>
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<h2>Algorithm</h2>
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<p>The implementation separately processes each segment that the
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base iterator returns.</p>
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<p>In the case of <code>SEG_CLOSE</code>,
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<code>SEG_MOVETO</code> and <code>SEG_LINETO</code> segments, the
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implementation simply hands the segment to the consumer, without actually
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doing anything.</p>
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<p>Any <code>SEG_QUADTO</code> and <code>SEG_CUBICTO</code> segments
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need to be flattened. Flattening is performed with a fixed-sized
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stack, holding the coordinates of subdivided segments. When the base
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iterator returns a <code>SEG_QUADTO</code> and
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<code>SEG_CUBICTO</code> segments, it is recursively flattened as
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follows:</p>
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<ol><li>Intialization: Allocate memory for the stack (unless a
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sufficiently large stack has been allocated previously). Push the
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original quadratic or cubic curve onto the stack. Mark that segment as
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having a <code>recLevel</code> of zero.</li>
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<li>If the stack is empty, flattening the segment is complete,
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and the next segment is fetched from the base iterator.</li>
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<li>If the stack is not empty, pop a curve segment from the
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stack.
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  <ul><li>If its <code>recLevel</code> exceeds the recursion limit,
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  hand the current segment to the consumer.</li>
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  <li>Calculate the squared flatness of the segment. If it smaller
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  than <code>flatnessSq</code>, hand the current segment to the
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  consumer.</li>
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  <li>Otherwise, split the segment in two halves. Push the right
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  half onto the stack. Then, push the left half onto the stack.
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  Continue with step two.</li></ul></li>
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</ol>
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<p>The implementation is slightly complicated by the fact that
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consumers <em>pull</em> the flattened segments from the
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<code>FlatteningPathIterator</code>. This means that we actually
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cannot “hand the curent segment over to the consumer.”
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But the algorithm is easier to understand if one assumes a
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<em>push</em> paradigm.</p>
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<h2>Example</h2>
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<p>The following example shows how a
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<code>FlatteningPathIterator</code> processes a
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<code>SEG_QUADTO</code> segment. It is (arbitrarily) assumed that the
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recursion limit was set to 2.</p>
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<blockquote>
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<table border="1" cellspacing="0" cellpadding="8">
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  <tr align="center" valign="baseline">
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    <th></th><th>A</th><th>B</th><th>C</th>
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    <th>D</th><th>E</th><th>F</th><th>G</th><th>H</th>
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  </tr>
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  <tr align="center" valign="baseline">
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    <th><code>stack[0]</code></th>
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    <td>—</td>
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    <td>—</td>
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    <td><i>S<sub>ll</sub>.x</i></td>
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    <td>—</td>
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    <td>—</td>
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    <td>—</td>
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    <td>—</td>
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    <td>—</td>
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  </tr>
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  <tr align="center" valign="baseline">
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    <th><code>stack[1]</code></th>
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    <td>—</td>
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    <td>—</td>
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    <td><i>S<sub>ll</sub>.y</i></td>
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    <td>—</td>
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    <td>—</td>
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    <td>—</td>
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    <td>—</td>
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    <td>—</td>
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  </tr>
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  <tr align="center" valign="baseline">
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    <th><code>stack[2]</code></th>
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    <td>—</td>
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    <td>—</td>
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    <td><i>C<sub>ll</sub>.x</i></td>
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    <td>—</td>
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    <td>—</td>
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    <td>—</td>
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    <td>—</td>
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    <td>—</td>
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  </tr>
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  <tr align="center" valign="baseline">
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    <th><code>stack[3]</code></th>
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    <td>—</td>
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    <td>—</td>
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    <td><i>C<sub>ll</sub>.y</i></td>
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    <td>—</td>
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    <td>—</td>
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    <td>—</td>
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    <td>—</td>
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    <td>—</td>
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  </tr>
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  <tr align="center" valign="baseline">
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    <th><code>stack[4]</code></th>
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    <td>—</td>
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    <td><i>S<sub>l</sub>.x</i></td>
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    <td><i>E<sub>ll</sub>.x</i>
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             = <i>S<sub>lr</sub>.x</i></td>
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    <td><i>S<sub>lr</sub>.x</i></td>
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    <td>—</td>
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    <td><i>S<sub>rl</sub>.x</i></td>
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    <td>—</td>
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    <td>—</td>
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  </tr>
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  <tr align="center" valign="baseline">
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    <th><code>stack[5]</code></th>
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    <td>—</td>
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    <td><i>S<sub>l</sub>.y</i></td>
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    <td><i>E<sub>ll</sub>.x</i>
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             = <i>S<sub>lr</sub>.y</i></td>
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    <td><i>S<sub>lr</sub>.y</i></td>
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    <td>—</td>
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    <td><i>S<sub>rl</sub>.y</i></td>
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    <td>—</td>
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    <td>—</td>
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  </tr>
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  <tr align="center" valign="baseline">
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    <th><code>stack[6]</code></th>
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    <td>—</td>
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    <td><i>C<sub>l</sub>.x</i></td>
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    <td><i>C<sub>lr</sub>.x</i></td>
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    <td><i>C<sub>lr</sub>.x</i></td>
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    <td>—</td>
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    <td><i>C<sub>rl</sub>.x</i></td>
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    <td>—</td>
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    <td>—</td>
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  </tr>
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  <tr align="center" valign="baseline">
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    <th><code>stack[7]</code></th>
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    <td>—</td>
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    <td><i>C<sub>l</sub>.y</i></td>
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    <td><i>C<sub>lr</sub>.y</i></td>
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    <td><i>C<sub>lr</sub>.y</i></td>
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    <td>—</td>
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    <td><i>C<sub>rl</sub>.y</i></td>
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    <td>—</td>
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    <td>—</td>
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  </tr>
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  <tr align="center" valign="baseline">
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    <th><code>stack[8]</code></th>
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    <td><i>S.x</i></td>
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    <td><i>E<sub>l</sub>.x</i>
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             = <i>S<sub>r</sub>.x</i></td>
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    <td><i>E<sub>lr</sub>.x</i>
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             = <i>S<sub>r</sub>.x</i></td>
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    <td><i>E<sub>lr</sub>.x</i>
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             = <i>S<sub>r</sub>.x</i></td>
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    <td><i>S<sub>r</sub>.x</i></td>
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    <td><i>E<sub>rl</sub>.x</i>
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             = <i>S<sub>rr</sub>.x</i></td>
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    <td><i>S<sub>rr</sub>.x</i></td>
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    <td>—</td>
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  </tr>
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  <tr align="center" valign="baseline">
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    <th><code>stack[9]</code></th>
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    <td><i>S.y</i></td>
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    <td><i>E<sub>l</sub>.y</i>
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             = <i>S<sub>r</sub>.y</i></td>
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    <td><i>E<sub>lr</sub>.y</i>
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             = <i>S<sub>r</sub>.y</i></td>
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    <td><i>E<sub>lr</sub>.y</i>
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             = <i>S<sub>r</sub>.y</i></td>
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    <td><i>S<sub>r</sub>.y</i></td>
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    <td><i>E<sub>rl</sub>.y</i>
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             = <i>S<sub>rr</sub>.y</i></td>
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    <td><i>S<sub>rr</sub>.y</i></td>
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    <td>—</td>
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  </tr>
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  <tr align="center" valign="baseline">
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    <th><code>stack[10]</code></th>
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    <td><i>C.x</i></td>
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    <td><i>C<sub>r</sub>.x</i></td>
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    <td><i>C<sub>r</sub>.x</i></td>
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    <td><i>C<sub>r</sub>.x</i></td>
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    <td><i>C<sub>r</sub>.x</i></td>
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    <td><i>C<sub>rr</sub>.x</i></td>
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    <td><i>C<sub>rr</sub>.x</i></td>
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    <td>—</td>
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  </tr>
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  <tr align="center" valign="baseline">
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    <th><code>stack[11]</code></th>
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    <td><i>C.y</i></td>
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    <td><i>C<sub>r</sub>.y</i></td>
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    <td><i>C<sub>r</sub>.y</i></td>
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    <td><i>C<sub>r</sub>.y</i></td>
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    <td><i>C<sub>r</sub>.y</i></td>
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    <td><i>C<sub>rr</sub>.y</i></td>
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    <td><i>C<sub>rr</sub>.y</i></td>
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    <td>—</td>
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  </tr>
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  <tr align="center" valign="baseline">
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    <th><code>stack[12]</code></th>
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    <td><i>E.x</i></td>
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    <td><i>E<sub>r</sub>.x</i></td>
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    <td><i>E<sub>r</sub>.x</i></td>
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    <td><i>E<sub>r</sub>.x</i></td>
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    <td><i>E<sub>r</sub>.x</i></td>
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    <td><i>E<sub>rr</sub>.x</i></td>
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    <td><i>E<sub>rr</sub>.x</i></td>
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    <td>—</td>
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  </tr>
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  <tr align="center" valign="baseline">
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    <th><code>stack[13]</code></th>
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    <td><i>E.y</i></td>
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    <td><i>E<sub>r</sub>.y</i></td>
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    <td><i>E<sub>r</sub>.y</i></td>
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    <td><i>E<sub>r</sub>.y</i></td>
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    <td><i>E<sub>r</sub>.y</i></td>
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    <td><i>E<sub>rr</sub>.y</i></td>
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    <td><i>E<sub>rr</sub>.x</i></td>
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    <td>—</td>
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  </tr>
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  <tr align="center" valign="baseline">
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     <th><code>stackSize</code></th>
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     <td>1</td>
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     <td>2</td>
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     <td>3</td>
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     <td>2</td>
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     <td>1</td>
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     <td>2</td>
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     <td>1</td>
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     <td>0</td>
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   </tr>
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   <tr align="center" valign="baseline">
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     <th><code>recLevel[2]</code></th>
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     <td>—</td>
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     <td>—</td>
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     <td>2</td>
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     <td>—</td>
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     <td>—</td>
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     <td>—</td>
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     <td>—</td>
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     <td>—</td>
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   </tr>
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   <tr align="center" valign="baseline">
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     <th><code>recLevel[1]</code></th>
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     <td>—</td>
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     <td>1</td>
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     <td>2</td>
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     <td>2</td>
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     <td>—</td>
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     <td>2</td>
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     <td>—</td>
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     <td>—</td>
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   </tr>
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   <tr align="center" valign="baseline">
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     <th><code>recLevel[0]</code></th>
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     <td>0</td>
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     <td>1</td>
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     <td>1</td>
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     <td>1</td>
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     <td>1</td>
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     <td>2</td>
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     <td>2</td>
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     <td>—</td>
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   </tr>
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 </table>
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</blockquote>
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<ol>
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<li>The data structures are initialized as follows.
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<ul><li>The segment’s end point <i>E</i>, control point
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<i>C</i>, and start point <i>S</i> are pushed onto the stack.</li>
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  <li>Currently, the curve in the stack would be approximated by one
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  single straight line segment (<i>S</i> – <i>E</i>).
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  Therefore, <code>stackSize</code> is set to 1.</li>
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  <li>This single straight line segment is approximating the original
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  curve, which can be seen as the result of zero recursive
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  splits. Therefore, <code>recLevel[0]</code> is set to
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  zero.</li></ul>
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Column A shows the state after the initialization step.</li>
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						|
   
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<li>The algorithm proceeds by taking the topmost curve segment
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(<i>S</i> – <i>C</i> – <i>E</i>) from the stack.
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						|
   
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  <ul><li>The recursion level of this segment (stored in
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  <code>recLevel[0]</code>) is zero, which is smaller than
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  the limit 2.</li>
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						|
   
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  <li>The method <code>java.awt.geom.QuadCurve2D.getFlatnessSq</code>
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  is called to calculate the squared flatness.</li>
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  <li>For the sake of argument, we assume that the squared flatness is
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  exceeding the threshold stored in <code>flatnessSq</code>. Thus, the
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  curve segment <i>S</i> – <i>C</i> – <i>E</i> gets
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  subdivided into a left and a right half, namely
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  <i>S<sub>l</sub></i> – <i>C<sub>l</sub></i> –
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  <i>E<sub>l</sub></i> and <i>S<sub>r</sub></i> –
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  <i>C<sub>r</sub></i> – <i>E<sub>r</sub></i>. Both halves are
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  pushed onto the stack, so the left half is now on top.
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  <br /> <br />The left half starts at the same point
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  as the original curve, so <i>S<sub>l</sub></i> has the same
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  coordinates as <i>S</i>.  Similarly, the end point of the right
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  half and of the original curve are identical
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  (<i>E<sub>r</sub></i> = <i>E</i>).  More interestingly, the left
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  half ends where the right half starts. Because
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  <i>E<sub>l</sub></i> = <i>S<sub>r</sub></i>, their coordinates need
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  to be stored only once, which amounts to saving 16 bytes (two
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						|
  <code>double</code> values) for each iteration.</li></ul>
 | 
						|
 | 
						|
Column B shows the state after the first iteration.</li>
 | 
						|
 | 
						|
<li>Again, the topmost curve segment (<i>S<sub>l</sub></i>
 | 
						|
– <i>C<sub>l</sub></i> – <i>E<sub>l</sub></i>) is
 | 
						|
taken from the stack.
 | 
						|
 | 
						|
  <ul><li>The recursion level of this segment (stored in
 | 
						|
  <code>recLevel[1]</code>) is 1, which is smaller than
 | 
						|
  the limit 2.</li>
 | 
						|
   
 | 
						|
  <li>The method <code>java.awt.geom.QuadCurve2D.getFlatnessSq</code>
 | 
						|
  is called to calculate the squared flatness.</li>
 | 
						|
 | 
						|
  <li>Assuming that the segment is still not considered
 | 
						|
  flat enough, it gets subdivided into a left
 | 
						|
  (<i>S<sub>ll</sub></i> – <i>C<sub>ll</sub></i> –
 | 
						|
  <i>E<sub>ll</sub></i>) and a right (<i>S<sub>lr</sub></i>
 | 
						|
  – <i>C<sub>lr</sub></i> – <i>E<sub>lr</sub></i>)
 | 
						|
  half.</li></ul>
 | 
						|
 | 
						|
Column C shows the state after the second iteration.</li>
 | 
						|
 
 | 
						|
<li>The topmost curve segment (<i>S<sub>ll</sub></i> –
 | 
						|
<i>C<sub>ll</sub></i> – <i>E<sub>ll</sub></i>) is popped from
 | 
						|
the stack.
 | 
						|
 | 
						|
  <ul><li>The recursion level of this segment (stored in
 | 
						|
  <code>recLevel[2]</code>) is 2, which is <em>not</em> smaller than
 | 
						|
  the limit 2. Therefore, a <code>SEG_LINETO</code> (from
 | 
						|
  <i>S<sub>ll</sub></i> to <i>E<sub>ll</sub></i>) is passed to the
 | 
						|
  consumer.</li></ul>
 | 
						|
 | 
						|
  The new state is shown in column D.</li>
 | 
						|
 | 
						|
 | 
						|
<li>The topmost curve segment (<i>S<sub>lr</sub></i> –
 | 
						|
<i>C<sub>lr</sub></i> – <i>E<sub>lr</sub></i>) is popped from
 | 
						|
the stack.
 | 
						|
 | 
						|
  <ul><li>The recursion level of this segment (stored in
 | 
						|
  <code>recLevel[1]</code>) is 2, which is <em>not</em> smaller than
 | 
						|
  the limit 2. Therefore, a <code>SEG_LINETO</code> (from
 | 
						|
  <i>S<sub>lr</sub></i> to <i>E<sub>lr</sub></i>) is passed to the
 | 
						|
  consumer.</li></ul>
 | 
						|
 | 
						|
  The new state is shown in column E.</li>
 | 
						|
 | 
						|
<li>The algorithm proceeds by taking the topmost curve segment
 | 
						|
(<i>S<sub>r</sub></i> – <i>C<sub>r</sub></i> –
 | 
						|
<i>E<sub>r</sub></i>) from the stack.
 | 
						|
   
 | 
						|
  <ul><li>The recursion level of this segment (stored in
 | 
						|
  <code>recLevel[0]</code>) is 1, which is smaller than
 | 
						|
  the limit 2.</li>
 | 
						|
  
 | 
						|
  <li>The method <code>java.awt.geom.QuadCurve2D.getFlatnessSq</code>
 | 
						|
  is called to calculate the squared flatness.</li>
 | 
						|
   
 | 
						|
  <li>For the sake of argument, we again assume that the squared
 | 
						|
  flatness is exceeding the threshold stored in
 | 
						|
  <code>flatnessSq</code>. Thus, the curve segment
 | 
						|
  (<i>S<sub>r</sub></i> – <i>C<sub>r</sub></i> –
 | 
						|
  <i>E<sub>r</sub></i>) is subdivided into a left and a right half,
 | 
						|
  namely
 | 
						|
  <i>S<sub>rl</sub></i> – <i>C<sub>rl</sub></i> –
 | 
						|
  <i>E<sub>rl</sub></i> and <i>S<sub>rr</sub></i> –
 | 
						|
  <i>C<sub>rr</sub></i> – <i>E<sub>rr</sub></i>. Both halves
 | 
						|
  are pushed onto the stack.</li></ul>
 | 
						|
 | 
						|
  The new state is shown in column F.</li>
 | 
						|
 | 
						|
<li>The topmost curve segment (<i>S<sub>rl</sub></i> –
 | 
						|
<i>C<sub>rl</sub></i> – <i>E<sub>rl</sub></i>) is popped from
 | 
						|
the stack.
 | 
						|
 | 
						|
  <ul><li>The recursion level of this segment (stored in
 | 
						|
  <code>recLevel[2]</code>) is 2, which is <em>not</em> smaller than
 | 
						|
  the limit 2. Therefore, a <code>SEG_LINETO</code> (from
 | 
						|
  <i>S<sub>rl</sub></i> to <i>E<sub>rl</sub></i>) is passed to the
 | 
						|
  consumer.</li></ul>
 | 
						|
 | 
						|
  The new state is shown in column G.</li>
 | 
						|
 | 
						|
<li>The topmost curve segment (<i>S<sub>rr</sub></i> –
 | 
						|
<i>C<sub>rr</sub></i> – <i>E<sub>rr</sub></i>) is popped from
 | 
						|
the stack.
 | 
						|
 | 
						|
  <ul><li>The recursion level of this segment (stored in
 | 
						|
  <code>recLevel[2]</code>) is 2, which is <em>not</em> smaller than
 | 
						|
  the limit 2. Therefore, a <code>SEG_LINETO</code> (from
 | 
						|
  <i>S<sub>rr</sub></i> to <i>E<sub>rr</sub></i>) is passed to the
 | 
						|
  consumer.</li></ul>
 | 
						|
 | 
						|
  The new state is shown in column H.</li>
 | 
						|
 | 
						|
<li>The stack is now empty. The FlatteningPathIterator will fetch the
 | 
						|
next segment from the base iterator, and process it.</li>
 | 
						|
 | 
						|
</ol>
 | 
						|
 | 
						|
<p>In order to split the most recently pushed segment, the
 | 
						|
<code>subdivideQuadratic()</code> method passes <code>stack</code>
 | 
						|
directly to
 | 
						|
<code>QuadCurve2D.subdivide(double[],int,double[],int,double[],int)</code>.
 | 
						|
Because the stack grows towards the beginning of the array, no data
 | 
						|
needs to be copied around: <code>subdivide</code> will directly store
 | 
						|
the result into the stack, which will have the contents shown to the
 | 
						|
right.</p>
 | 
						|
 | 
						|
</body>
 | 
						|
</html>
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