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  • opengl paint program based on Apple's 'glPaint' on a white background - how to blend?

    - by Adam
    Trying to write a simple paint program for iPhone, and I'm using Apple's glPaint sample as a guide. The only problem is, painting doesn't work on a white background, since white + colour = white. I've tried different blending functions, but haven't been able to hit on the right combination of settings and/or brushes to make this work. I've seen similar posts about this problem but no answers. Does anyone know how this might work?

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  • How do you implement position-sensitive zooming inside a JScrollPane?

    - by tucuxi
    I am trying to implement position-sensitive zooming inside a JScrollPane. The JScrollPane contains a component with a customized 'paint' that will draw itself inside whatever space it is allocated - so zooming is as easy as using a MouseWheelListener that resizes the inner component as required. But I also want zooming into (or out of) a point to keep that point as central as possible within the resulting zoomed-in (or -out) view (this is what I refer to as 'position-sensitive' zooming), similar to how zooming works in google maps. I am sure this has been done many times before - does anybody know the "right" way to do it under Java Swing?. Would it be better to play with Graphic2D's transformations instead of using JScrollPanes? Sample code follows: package test; import java.awt.*; import java.awt.event.*; import java.awt.geom.*; import javax.swing.*; public class FPanel extends javax.swing.JPanel { private Dimension preferredSize = new Dimension(400, 400); private Rectangle2D[] rects = new Rectangle2D[50]; public static void main(String[] args) { JFrame jf = new JFrame("test"); jf.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); jf.setSize(400, 400); jf.add(new JScrollPane(new FPanel())); jf.setVisible(true); } public FPanel() { // generate rectangles with pseudo-random coords for (int i=0; i<rects.length; i++) { rects[i] = new Rectangle2D.Double( Math.random()*.8, Math.random()*.8, Math.random()*.2, Math.random()*.2); } // mouse listener to detect scrollwheel events addMouseWheelListener(new MouseWheelListener() { public void mouseWheelMoved(MouseWheelEvent e) { updatePreferredSize(e.getWheelRotation(), e.getPoint()); } }); } private void updatePreferredSize(int n, Point p) { double d = (double) n * 1.08; d = (n > 0) ? 1 / d : -d; int w = (int) (getWidth() * d); int h = (int) (getHeight() * d); preferredSize.setSize(w, h); getParent().doLayout(); // Question: how do I keep 'p' centered in the resulting view? } public Dimension getPreferredSize() { return preferredSize; } private Rectangle2D r = new Rectangle2D.Float(); public void paint(Graphics g) { super.paint(g); g.setColor(Color.red); int w = getWidth(); int h = getHeight(); for (Rectangle2D rect : rects) { r.setRect(rect.getX() * w, rect.getY() * h, rect.getWidth() * w, rect.getHeight() * h); ((Graphics2D)g).draw(r); } } }

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  • Is the video recording possible in the background service?

    - by pyt
    Hello, All. I've tried video recording in the background. But had failed. Under normal circumstances, the recording works properly. However, if HOME key down- Home screen or Other Activity is running, recording terminates. In such a situation, I want to record continuously. I want to record whole process! What should I do? thanks. -Michael

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  • Finding N contiguous zero bits in an integer to the left of the MSB position of another integer

    - by James Morris
    The problem is: given an integer val1 find the position of the highest bit set (Most Significant Bit) then, given a second integer val2 find a contiguous region of unset bits, with the minimum number of zero bits given by width to the left of the position (ie, in the higher bits). Here is the C code for my solution: typedef unsigned int t; unsigned const t_bits = sizeof(t) * CHAR_BIT; _Bool test_fit_within_left_of_msb( unsigned width, t val1, t val2, unsigned* offset_result) { unsigned offbit = 0; unsigned msb = 0; t mask; t b; while(val1 >>= 1) ++msb; while(offbit + width < t_bits - msb) { mask = (((t)1 << width) - 1) << (t_bits - width - offbit); b = val2 & mask; if (!b) { *offset_result = offbit; return true; } if (offbit++) /* this conditional bothers me! */ b <<= offbit - 1; while(b <<= 1) offbit++; } return false; } Aside from faster ways of finding the MSB of the first integer, the commented test for a zero offbit seems a bit extraneous, but necessary to skip the highest bit of type t if it is set. I have also implemented similar algorithms but working to the right of the MSB of the first number, so they don't require this seemingly extra condition. How can I get rid of this extra condition, or even, are there far more optimal solutions? Edit: Some background not strictly required. The offset result is a count of bits from the high bit, not from the low bit as maybe expected. This will be part of a wider algorithm which scans a 2D array for a 2D area of zero bits. Here, for testing, the algorithm has been simplified. val1 represents the first integer which does not have all bits set found in a row of the 2D array. From this the 2D version would scan down which is what val2 represents. Here's some output showing success and failure: t_bits:32 t_high: 10000000000000000000000000000000 ( 2147483648 ) --------- ----------------------------------- *** fit within left of msb test *** ----------------------------------- val1: 00000000000000000000000010000000 ( 128 ) val2: 01000001000100000000100100001001 ( 1091569929 ) msb: 7 offbit:0 + width: 8 = 8 mask: 11111111000000000000000000000000 ( 4278190080 ) b: 01000001000000000000000000000000 ( 1090519040 ) offbit:8 + width: 8 = 16 mask: 00000000111111110000000000000000 ( 16711680 ) b: 00000000000100000000000000000000 ( 1048576 ) offbit:12 + width: 8 = 20 mask: 00000000000011111111000000000000 ( 1044480 ) b: 00000000000000000000000000000000 ( 0 ) offbit:12 iters:10 ***** found room for width:8 at offset: 12 ***** ----------------------------------- *** fit within left of msb test *** ----------------------------------- val1: 00000000000000000000000001000000 ( 64 ) val2: 00010000000000001000010001000001 ( 268469313 ) msb: 6 offbit:0 + width: 13 = 13 mask: 11111111111110000000000000000000 ( 4294443008 ) b: 00010000000000000000000000000000 ( 268435456 ) offbit:4 + width: 13 = 17 mask: 00001111111111111000000000000000 ( 268402688 ) b: 00000000000000001000000000000000 ( 32768 ) ***** mask: 00001111111111111000000000000000 ( 268402688 ) offbit:17 iters:15 ***** no room found for width:13 ***** (iters is the count of iterations of the inner while loop)

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  • What happens on Activity.finish() with AsyncTask still running in background?

    - by stormin986
    What happens on Activity.finish() with an AsyncTask still running in background? Does it just pop the Activity off the Activity Stack, but wait to destroy the Activity object until the AsyncTask fully completes (since the AsyncTask is an inner class of my Activity)? Also, would it act any differently if the AsyncTask were a public, non-inner class that held no references to the instance of the Activity?

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