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PORemoval of container which uses a complex layout
 

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  1. PORemoval of container which uses a complex layout
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    <p>I'm creating an (j)applet where the user has access to several different views selectable from a tree. The applet has a main area where the current view is displayed inside a scrollpane, replacing whatever was previously there. Most of the time this works fine, but when the view to be replaced has a more complicated layout, as in large (~100 components), not nested, the removal takes several seconds! Creation of the large view is lightning quick, only the removal is slow. This is only noticeable in applets, running the application in a JFrame is still as fast as I expect. I use GridBagLayout for the view.</p> <p>After some experimentation, I found that if I replace</p> <pre><code>contents.remove(view); //contents is the view of the scrollpane. </code></pre> <p>with</p> <pre><code>view.removeAll(); contents.remove(view); </code></pre> <p>then the whole process is very fast.</p> <p>My guess is that the cause of this behavior is some of Java's layout "magic", and although I have a solution it doesn't feel right and seems to indicate that I don't fully understand what's going on. So my question(s) is this:<br> Does someone know what's actually happening? Is there a better (clearer) way to accomplish the same thing?</p> <p>Edit: SSCCE: <a href="http://pastebin.com/VC1T7zGv" rel="nofollow">http://pastebin.com/VC1T7zGv</a></p> <pre><code>import java.awt.BorderLayout; import java.awt.GridBagConstraints; import java.awt.GridBagLayout; import java.awt.event.ActionEvent; import javax.swing.AbstractAction; import javax.swing.JApplet; import javax.swing.JButton; import javax.swing.JLabel; import javax.swing.JPanel; public class SSCCE extends JApplet { private boolean state; public void init() { super.init(); getContentPane().add(setupSwitchPanel(), BorderLayout.NORTH); getContentPane().add(setupSimplePanel(), BorderLayout.CENTER); } private JPanel setupSwitchPanel() { JPanel switchPanel = new JPanel(); switchPanel.add(new JButton(new AbstractAction("Switch panels (slow)") { @Override public void actionPerformed(ActionEvent arg0) { switchComponents(); } })); switchPanel.add(new JButton(new AbstractAction("Switch panels (fast)") { @Override public void actionPerformed(ActionEvent arg0) { ((JPanel)getContentPane().getComponent(1)).removeAll(); switchComponents(); } })); return switchPanel; } private void switchComponents() { getContentPane().remove(1); if (state) getContentPane().add(setupSimplePanel(), BorderLayout.CENTER); else getContentPane().add(setupComplexPanel(), BorderLayout.CENTER); state = !state; getContentPane().validate(); } private JPanel setupComplexPanel() { JPanel contents = new JPanel(); contents.setLayout(new GridBagLayout()); GridBagConstraints c = new GridBagConstraints(); for (int x = 0; x&lt;10; x++) { c.gridx = x; for (int y = 0; y&lt;10; y++) { c.gridy = y; contents.add(new JLabel(x + ", " + y), c); } } return contents; } private JPanel setupSimplePanel() { JPanel contents = new JPanel(); contents.add(new JLabel("Simple view")); return contents; } } </code></pre>
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    1. USAndrew Thompson
    1. USJacob Raihle
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    1. CO@AndrewThompson A CardLayout might work if I put the views in separate JScrollPanes, but I would prefer to not keep all of the different views in memory.
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      1. USJacob Raihle
    2. COI decided to give it a try and this might work - I was actually expecting to get the same behavior. CardLayout may well be the most idiomatic solution as well. I'll probably go with this, but I'm still curious as to why removal of sub-components makes removal of the parent component quicker.
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      1. PORemoval of container which uses a complex layout
      1. USJacob Raihle
    3. CO1) I entered the comments as an alternate strategy. 2) Nice SSCCE +1 - I pasted it into the question as an edit. 3) Intriguing observation, but I'd tend to use the best layout & move on to some new challenge. Hope you get a solution. :)
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      1. USAndrew Thompson
 

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