Tag Archives: flash

AS Quiz # 2

By Zevan | January 12, 2010

People seemed to do pretty well on yesterday’s quiz. I have lots of questions written that fall into that same category of being somewhere between beginner and intermediate level. I decided to write a trickier quiz for today. Tomorrow I’ll alternate back to some easier questions. This quiz is really more about operators and hex numbers than ActionScript…

Number of Questions : 5
Difficulty : Medium
Topic : Operators and Hexadecimal

Assuming that we're using hexadecimal to represent colors (as in the case of beginFill() or lineStyle())...
Which of the following hexadecimal numbers has the highest value in its green channel?





Which answer best describes the following operators:
& | ! % ===






Which of the following would correctly unpack the red channel from a hexadecimal color.





Which of the following operators would you most likely use for integer multiplication?






Which of the below is equivalent to the following expression?

110 % 3







Posted in Quiz | Also tagged , , | 14 Comments

AS Quiz #1

By Zevan | January 11, 2010

This is the first in a hopefully long series of short ActionScript Quizzes I’ll be writing. This is just 6 simple questions… check it out….

Number of Questions : 6
Difficulty : Easy
Topic : General

Please go to AS Quiz #1 to view the quiz

Posted in Quiz | Also tagged , , | 16 Comments

Too Many Buttons

By Zevan | January 4, 2010
CLICK HERE TO COPY
Actionscript:
  1. [SWF (width = 500, height = 500)]
  2.  
  3. var canvas:BitmapData = new BitmapData(stage.stageWidth, stage.stageHeight, false, 0xFFFFFF);
  4. addChild(new Bitmap(canvas));
  5.  
  6. var indexCanvas:BitmapData = new BitmapData(stage.stageWidth, stage.stageHeight, false, 0xFFFFFF);
  7.  
  8. var btnNum:int = 5000;
  9. var info:Array = [];
  10.  
  11. var brush:BitmapData = new BitmapData(10,10,false, 0xCCCCCC);
  12. var border:Shape = new Shape();
  13. border.graphics.lineStyle(2, 0x000000);
  14. border.graphics.drawRect(0,0,10,10);
  15. brush.draw(border);
  16.  
  17. var txt:TextField = TextField(addChild(new TextField()));
  18. with (txt) height = 20, width = 50, background = 0xFFFFFF, selectable = false
  19. var tf:TextFormat = new TextFormat();
  20. tf.align = TextFormatAlign.RIGHT;
  21. txt.border= true;
  22. txt.defaultTextFormat = tf;
  23.  
  24. var redRect:Shape = Shape(addChild(new Shape()));
  25. with (redRect.graphics) beginFill(0xFF0000), drawRect(0,0,10,10);
  26.                                              
  27. var pnt:Point = new Point();
  28. var r:Rectangle = new Rectangle(0,0,10,10);
  29. for (var i:int = 0; i <btnNum; i++){
  30.     pnt.x = r.x = int(Math.random() * stage.stageWidth);
  31.     pnt.y = r.y = int(Math.random() * stage.stageHeight);
  32.     indexCanvas.fillRect(r, i);
  33.     canvas.copyPixels(brush, brush.rect, pnt)
  34.     info[i] = [r.x, r.y, i];   
  35. }
  36.  
  37. addEventListener(Event.ENTER_FRAME, onCheckBtns);
  38. function onCheckBtns(evt:Event):void{
  39.    var currentIndex:int = indexCanvas.getPixel(mouseX, mouseY);
  40.    if (currentIndex != 0xFFFFFF){
  41.      var currentBox:Array = info[currentIndex]
  42.      redRect.visible = true;
  43.      redRect.x = currentBox[0];
  44.      txt.y = redRect.y = currentBox[1];
  45.      if (mouseX <txt.width){
  46.          tf.align = TextFormatAlign.LEFT;
  47.          txt.defaultTextFormat = tf;
  48.          txt.x = redRect.x + 10;
  49.      }else{
  50.          tf.align = TextFormatAlign.RIGHT;
  51.          txt.defaultTextFormat = tf;
  52.          txt.x = redRect.x - txt.width;
  53.      }
  54.      txt.text = currentBox[2];
  55.      txt.visible = true;
  56.    }else{
  57.      redRect.visible = false;
  58.      txt.visible = false;
  59.    }
  60. }

This is a simplified example of the technique discussed in yesterdays post. The idea is to use a BitmapData image to store index values for a large number of elements that need to be able to act as if the have MouseEvents. For a more detailed description of this technique see yesterdays post.

Have a look at the swf on wonderfl

Posted in BitmapData, arrays, misc | Also tagged , | Leave a comment

~20,000 Rollovers

By Zevan | January 3, 2010
CLICK HERE TO COPY
Actionscript:
  1. [SWF(width = 500, height = 500, frameRate = 30)]
  2.  
  3. var canvas:BitmapData = new BitmapData(stage.stageWidth,stage.stageHeight,false, 0xFFFFFF);
  4.  
  5. var indexCanvas:BitmapData = new BitmapData(stage.stage.stageWidth, stage.stageHeight, false,
  6.                                             0xFFFFFF);
  7. addChild(new Bitmap(canvas));
  8.  
  9. var s:Shape = new Shape();
  10.  
  11. var lineData:Array = [];
  12. var dataIndex:int = 0;
  13.  
  14. trace(0xFFFFFF - 1)
  15. var totalLines:int = 20000;
  16. var iterations:int = 9;
  17. var linesPerIter:int = totalLines / iterations;
  18.  
  19. var xp:int = stage.stageWidth / 2;
  20. var yp:int = stage.stageHeight / 2;
  21.  
  22. var stepAmt:Number = 60;
  23. var halfStepAmt:Number = stepAmt / 2;
  24.  
  25. addEventListener(Event.ENTER_FRAME, onDraw);
  26. function onDraw(evt:Event):void {
  27.      if (lineData.length <totalLines){
  28.         generateData(linesPerIter);
  29.      }else{
  30.         stage.quality = "high";
  31.         addChild(s);
  32.         s.x = 0;
  33.         s.y = 0;
  34.          
  35.         removeEventListener(Event.ENTER_FRAME, onDraw);
  36.         addEventListener(Event.ENTER_FRAME, onRun);
  37.      }
  38. }
  39.  
  40. function onRun(evt:Event):void {
  41.    var currentIndex:int = indexCanvas.getPixel(mouseX, mouseY);
  42.    var currentLine:Array = lineData[currentIndex];
  43.    
  44.    s.graphics.clear();
  45.    if (currentIndex != 0xFFFFFF){
  46.           s.graphics.lineStyle(3, 0xFF0000);
  47.           s.graphics.moveTo(currentLine[0], currentLine[1]);
  48.           s.graphics.lineTo(currentLine[2], currentLine[3]);  
  49.    }
  50. }
  51.  
  52. function generateData(num:int):void{
  53.     var rxA:int, rxB:int, ryA:int, ryB:int;
  54.     var g:Graphics = s.graphics;
  55.     for (var i:int = 0; i<num; i++){
  56.         rxA = xp;
  57.         ryA = yp;
  58.        
  59.         xp += Math.round(Math.random() * stepAmt) - halfStepAmt;
  60.         yp += Math.round(Math.random() * stepAmt) - halfStepAmt;
  61.        
  62.         if (xp> stage.stageWidth){
  63.             xp = stage.stageWidth - halfStepAmt;
  64.         }else
  65.         if (xp <0){
  66.             xp = halfStepAmt;
  67.         }
  68.         if (yp> stage.stageHeight){
  69.             yp = stage.stageHeight - halfStepAmt;
  70.         }else
  71.         if (yp <0){
  72.             yp = halfStepAmt;
  73.         }
  74.        
  75.         rxB = xp;
  76.         ryB = yp;
  77.          
  78.         lineData[dataIndex] = [rxA, ryA, rxB, ryB];            
  79.         s.x = rxA;
  80.         s.y = ryA;
  81.         var endX:Number = rxB - rxA;
  82.         var endY:Number = ryB - ryA;
  83.         var m:Matrix = s.transform.matrix;
  84.         g.clear();
  85.         g.lineStyle(1, 0x000000, 0.3);
  86.  
  87.         g.lineTo(endX, endY);
  88.         stage.quality = "high";
  89.         canvas.draw(s, m);
  90.        
  91.         g.clear();
  92.         g.lineStyle(3, dataIndex);
  93.        
  94.         g.lineTo(endX, endY);
  95.         stage.quality = "low";
  96.         indexCanvas.draw(s, m);
  97.        
  98.         dataIndex++
  99.     }
  100. }

I'm working on a data visualization that contains a long path made up of approximately one million points. There is some information associated with every two sets of coordinates that needs to be displayed when the user rolls their mouse over any part of the line.

I took a little time to think about the best way to do this and came up with a few techniques. The first one I tried seems to work nicely - this snippet is the proof of concept for that first technique. I tested this snippet with 1,000,000 xy coordinates and it works nicely. It takes a little while to draw though, so for the purposes of this demo I've just included 20,000 coordinates.

Have a look at the swf over at wonderfl.net

The way this works is by drawing lines to two different BitmapData instances. I draw anti-aliased slightly transparent lines to a BitmapData instance called "canvas" (this is added to the display list) - I then draw aliased lines to a BitmapData called "indexCanvas" (this is never added to the display list) - each aliased line uses an incremental value for its color - this incremental value is also the index for a two dimensional array containing the coordinate information for the aliased line. I use getPixel() on the "indexCanvas" and use the return value as the index for the 2D array. The data from the 2D array is used to draw a red line with the graphics class. This technique enables you to have many many rollovers and all you ever have to do is call getPixel() and use the returned color value to look up info about what you're mouse is touching.

There are a few cool ways this could be repurposed and this is really only one solution to the problem of having many many things that you need to be able to rollover... there are others that don't use BitmapData at all... I may write those up in the next couple of days.

Posted in BitmapData, Data Structures, UI, arrays, display list, graphics algorithms, matrix, misc, pixel manipulation, return values | Also tagged , | 2 Comments