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To be more precise, the part of the NXT firmware termed scheduler is responsible for this task execution switching Multitasking is also called concurrent programming, if the tasks in execution are sharing some resources, such as a sensor, an output device, a motor, or a display This resource sharing must be well disciplined to avoid weird and unpredictable behavior in the program; this is the main trap of multitasking As an example to make this clearer, imagine a narrow kitchen of a restaurant, equipped with pots, ladles, but only one oven If only one cook is at work, there s no problem: he can use all the equipment with ease However, if the restaurant owner decides to improve his kitchen, engaging one more cook and a chef, then things change a lot. barcode reader using java source code Barcode Scanner implementation on Java - Stack Overflow
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Free source code and tutorials for Software developers and Architects.; Updated: 10 Jun 2015. // you check myHeldKey for null because you don't // want to accidentally pick up two keys. if((myPrincess.collidesWith(myKeys[i], true)) && (myHeldKey == null)) { myHeldKey = myKeys[i]; myHeldKey.setVisible(false); } } if(oldHeld != null) { // place the key you're putting down in the princess's // current position and make it visible: oldHeld.setPosition(myPrincess.getX(), myPrincess.getY()); oldHeld.setVisible(true); } } } /** * Checks if the player hits a stone wall or a door. */ boolean checkCollision() { boolean retVal = false; // the "true" arg means to check for a pixel-level // collision (so merely an overlap in image // squares does not register as a collision) if(myPrincess.collidesWith(myBackground, true)) { retVal = true; } else { // Note: it is not necessary to synchronize // this block because the thread that calls this // method is the same as the one that puts down the // keys, so there's no danger of the key being put down // between the moment you check for the key and // the moment you open the door: for(int i = 0; i < myDoors.length; i++) { // if she's holding the right key, then open the door // otherwise bounce off if(myPrincess.collidesWith(myDoors[i], true)) { if((myHeldKey != null) && (myDoors[i].getColor() == myHeldKey.getColor())) { setNeedsRepaint(); myDoors[i].setVisible(false); } else { // if she's not holding the right key, then // she has collided with the door just the same // as if she had collided with a wall: retVal = true; } barcode scanner code in java Read USB Barcode Scanner Data in Java - Stack Overflow
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