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  • Why does creating dynamic bodies in JBox2D freeze my app?

    - by Amplify91
    My game hangs/freezes when I create dynamic bullet objects with Box2D and I don't know why. I am making a game where the main character can shoot bullets by the user tapping on the screen. Each touch event spawns a new FireProjectileEvent that is handled properly by an event queue. So I know my problem is not trying to create a new body while the box2d world is locked. My bullets are then created and managed by an object pool class like this: public Projectile getProjectile(){ for(int i=0;i<mProjectiles.size();i++){ if(!mProjectiles.get(i).isActive){ return mProjectiles.get(i); } } return mSpriteFactory.createProjectile(); } mSpriteFactory.createProjectile() leads to the physics component of the Projectile class creating its box2d body. I have narrowed the issue down to this method and it looks like this: public void create(World world, float x, float y, Vec2 vertices[], boolean dynamic){ BodyDef bodyDef = new BodyDef(); if(dynamic){ bodyDef.type = BodyType.DYNAMIC; }else{ bodyDef.type = BodyType.STATIC; } bodyDef.position.set(x, y); mBody = world.createBody(bodyDef); PolygonShape dynamicBox = new PolygonShape(); dynamicBox.set(vertices, vertices.length); FixtureDef fixtureDef = new FixtureDef(); fixtureDef.shape = dynamicBox; fixtureDef.density = 1.0f; fixtureDef.friction = 0.0f; mBody.createFixture(fixtureDef); mBody.setFixedRotation(true); } If the dynamic parameter is set to true my game freezes before crashing, but if it is false, it will create a projectile exactly how I want it just doesn't function properly (because a projectile is not a static object). Why does my program fail when I try to create a dynamic object at runtime but not when I create a static one? I have other dynamic objects (like my main character) that work fine. Any help would be greatly appreciated. This is a screenshot of a method profile I did: Especially notable is number 8. I'm just still unsure what I'm doing wrong. Other notes: I am using JBox2D 2.1.2.2. (Upgraded from 2.1.2.1 to try to fix this problem) When the application freezes, if I hit the back button, it appears to move my game backwards by one update tick. Very strange.

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  • Air Canada Will No Longer Be My Airline

    - by D'Arcy Lussier
    If the constant labour disputes at Air Canada (the most recent being a week ago where pilots were locked out and mechanics and bag handlers were poised to strike) weren’t enough to make me reconsider moving all my flights to West Jet, this latest twist definitely will. CBC reported that Aveos, a privately held company that has the contract to provide maintenance for Air Canada, had suddenly and without notice shut its doors (read the story here) There’s something missing from the stories currently online though. Months ago, Air Canada gave their Winnipeg based maintenance staff an ultimatum – stay with Air Canada but be forced to relocate to a different city, or switch from Air Canada to Aveos and stay in Winnipeg. So all of those staff that Air Canada pushed into Aveos just so they could stay in Winnipeg are now out of a job with huge uncertainty around their future. Labour disputes that rise up continually and hamper personal travel and business, questionable timing of business decisions and the resulting impact on individuals…there’s too much drama in that company for me to rely on it for my travel needs. WestJet it is moving forward until Air Canada gets their act together – which probably means its WestJet for the foreseeable future. D

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  • Why does my code dividing a 2D array into chunks fail?

    - by Borog
    I have a 2D-Array representing my world. I want to divide this huge thing into smaller chunks to make collision detection easier. I have a Chunk class that consists only of another 2D Array with a specific width and height and I want to iterate through the world, create new Chunks and add them to a list (or maybe a Map with Coordinates as the key; we'll see about that). world = new World(8192, 1024); Integer[][] chunkArray; for(int a = 0; a < map.getHeight() / Chunk.chunkHeight; a++) { for(int b = 0; b < map.getWidth() / Chunk.chunkWidth; b++) { Chunk chunk = new Chunk(); chunkArray = new Integer[Chunk.chunkWidth][Chunk.chunkHeight]; for(int x = Chunk.chunkHeight*a; x < Chunk.chunkHeight*(a+1); x++) { for(int y = Chunk.chunkWidth*b; y < Chunk.chunkWidth*(b+1); y++) { // Yes, the tileMap actually is [height][width] I'll have // to fix that somewhere down the line -.- chunkArray[y][x] = map.getTileMap()[x*a][y*b]; // TODO:Attach to chunk } } chunkList.add(chunk); } } System.out.println(chunkList.size()); The two outer loops get a new chunk in a specific row and column. I do that by dividing the overall size of the map by the chunkSize. The inner loops then fill a new chunkArray and attach it to the chunk. But somehow my maths is broken here. Let's assume the chunkHeight = chunkWidth = 64. For the first Array I want to start at [0][0] and go until [63][63]. For the next I want to start at [64][64] and go until [127][127] and so on. But I get an out of bounds exception and can't figure out why. Any help appreciated! Actually I think I know where the problem lies: chunkArray[y][x] can't work, because y goes from 0-63 just in the first iteration. Afterwards it goes from 64-127, so sure it is out of bounds. Still no nice solution though :/ EDIT: if(y < Chunk.chunkWidth && x < Chunk.chunkHeight) chunkArray[y][x] = map.getTileMap()[y][x]; This works for the first iteration... now I need to get the commonly accepted formula.

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  • Congratulations to 2012 Innovation Award winners in BPM category

    - by Manoj Das
    Last year many of our customers went live on BPM 11g. It is my extreme pleasure to congratulate two of them – Amadeus and Navistar – for being awarded Oracle Fusion Middleware Innovation Award at Oracle OpenWorld 2012. We invited our customers to submit their most innovative BPM implementations that have delivered substantiated value to them. This year we saw more than 20 submissions from our customers seeing significant business value from their live BPM 11g deployments. The submissions came from across the world, spanning various industry verticals including manufacturing, healthcare, logistics, Hi-Tech, Public Sector, Education and covering many process usage patterns. Award submissions were evaluated based on the uniqueness of their business case, business benefits, level of impact relative to the size of the organization, complexity and magnitude of implementation, and the originality of architecture. Amadeus Team Receiving Innovation Award from Hasan Rizvi Congratulations to Amadeus and Navistar and their teams on being recognized from among some very strong submissions and more importantly for the business value delivered. It is an honor to be part of your success and to play a small role in the innovation you drive. Navistar is a leading truck manufacturing company which produces International® brand commercial and military trucks, MaxxForce® brand diesel engines, IC Bus™ brand school and commercial buses, and Navistar RV brands of recreational vehicles. The company also provides truck and diesel engine service parts. Amadeus is a leading transaction processor for the global travel and tourism industry, providing transaction processing power and technology solutions to both travellers and travel providers. Both Navistar and Amadeus have leveraged Oracle BPM Suite to improve visibility into their business and made their business more agile and efficient. We congratulate them again and wish them continued success in their business and future BPM initiatives.

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  • Leeroy Jenkins Reimagined As a Bank Heist [Video]

    - by Jason Fitzpatrick
    In this 3-minute short film, Leeroy Jenkins of World of War Craft infamy, brings his raid-botching power to the real world in a bank heist gone wrong. If you haven’t seen the original Leeroy Jenkins clip, watch it first to properly contextualize this entertaining little film. [via Super Punch] HTG Explains: Learn How Websites Are Tracking You Online Here’s How to Download Windows 8 Release Preview Right Now HTG Explains: Why Linux Doesn’t Need Defragmenting

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  • Certify May Updates

    - by Sadia2
    We have added some release and platform certifications to MOS Certify Database: Oracle Database 10.2.0.5.0, Oracle Fail Safe Server 4.1.0 Fusion Middleware: Oracle Tuxedo 10.3.0.0.0, Oracle Business Intelligence Applications 11.1.1.7.1 E-Business Suite: Oracle E-Business Suite 12.1.2 JD Edwards EnterpriseOne: JD Edwards EnterpriseOne Database Server for In-Memory Applications X9.1.3.0, JD Edwards EnterpriseOne Business Services Server 9.1.3.0 JD Edwards World: JD Edwards World Base product A9.3-Single Byte  

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  • Finding the shortest path through a digraph that visits all nodes

    - by Boluc Papuccuoglu
    I am trying to find the shortest possible path that visits every node through a graph (a node may be visited more than once, the solution may pick any node as the starting node.). The graph is directed, meaning that being able to travel from node A to node B does not mean one can travel from node B to node A. All distances between nodes are equal. I was able to code a brute force search that found a path of only 27 nodes when I had 27 nodes and each node had a connection to 2 or 1 other node. However, the actual problem that I am trying to solve consists of 256 nodes, with each node connecting to either 4 or 3 other nodes. The brute force algorithm that solved the 27 node graph can produce a 415 node solution (not optimal) within a few seconds, but using the processing power I have at my disposal takes about 6 hours to arrive at a 402 node solution. What approach should I use to arrive at a solution that I can be certain is the optimal one? For example, use an optimizer algorithm to shorten a non-optimal solution? Or somehow adopt a brute force search that discards paths that are not optimal? EDIT: (Copying a comment to an answer here to better clarify the question) To clarify, I am not saying that there is a Hamiltonian path and I need to find it, I am trying to find the shortest path in the 256 node graph that visits each node AT LEAST once. With the 27 node run, I was able to find a Hamiltonian path, which assured me that it was an optimal solution. I want to find a solution for the 256 node graph which is the shortest.

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  • Box2D Difference Between WorldCenter and Position

    - by Free Lancer
    So this problem has been brothering for a couple of days now. First off, what is the difference between say Body.getWorldCenter() and Body.getPosition(). I heard that WorldCenter might have to do with the center of gravity or something. Second, When I create a Box2D Body for a sprite the Body is always at the lower left corner. I check it by printing a Rectangle of 1 pixel around the box.getWorldCenter(). From what I understand the Body should be in the center of the Sprite and its bounding box should wrap around the Sprite, correct? Here's an image of what I mean (The Sprite is Red, Body Blue): Here's some code: Body Creator: public static Body createBoxBody( final World pPhysicsWorld, final BodyType pBodyType, final FixtureDef pFixtureDef, Sprite pSprite ) { float pRotation = 0; float pCenterX = pSprite.getX() + pSprite.getWidth() / 2; float pCenterY = pSprite.getY() + pSprite.getHeight() / 2; float pWidth = pSprite.getWidth(); float pHeight = pSprite.getHeight(); final BodyDef boxBodyDef = new BodyDef(); boxBodyDef.type = pBodyType; //boxBodyDef.position.x = pCenterX / Constants.PIXEL_METER_RATIO; //boxBodyDef.position.y = pCenterY / Constants.PIXEL_METER_RATIO; boxBodyDef.position.x = pSprite.getX() / Constants.PIXEL_METER_RATIO; boxBodyDef.position.y = pSprite.getY() / Constants.PIXEL_METER_RATIO; Vector2 v = new Vector2( boxBodyDef.position.x * Constants.PIXEL_METER_RATIO, boxBodyDef.position.y * Constants.PIXEL_METER_RATIO ); Gdx.app.log("@Physics", "createBoxBody():: Box Position: " + v); // Temporary Box shape of the Body final PolygonShape boxPoly = new PolygonShape(); final float halfWidth = pWidth * 0.5f / Constants.PIXEL_METER_RATIO; final float halfHeight = pHeight * 0.5f / Constants.PIXEL_METER_RATIO; boxPoly.setAsBox( halfWidth, halfHeight ); // set the anchor point to be the center of the sprite pFixtureDef.shape = boxPoly; final Body boxBody = pPhysicsWorld.createBody(boxBodyDef); Gdx.app.log("@Physics", "createBoxBody():: Box Center: " + boxBody.getPosition().mul(Constants.PIXEL_METER_RATIO)); boxBody.createFixture(pFixtureDef); boxBody.setTransform( boxBody.getWorldCenter(), MathUtils.degreesToRadians * pRotation ); boxPoly.dispose(); return boxBody; } Making the Sprite: public Car( Texture texture, float pX, float pY, World world ) { super( "Car" ); mSprite = new Sprite( texture ); mSprite.setSize( mSprite.getWidth() / 6, mSprite.getHeight() / 6 ); mSprite.setPosition( pX, pY ); mSprite.setOrigin( mSprite.getWidth()/2, mSprite.getHeight()/2); FixtureDef carFixtureDef = new FixtureDef(); // Set the Fixture's properties, like friction, using the car's shape carFixtureDef.restitution = 1f; carFixtureDef.friction = 1f; carFixtureDef.density = 1f; // needed to rotate body using applyTorque mBody = Physics.createBoxBody( world, BodyDef.BodyType.DynamicBody, carFixtureDef, mSprite ); }

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  • Parallel task in C# 4.0

    - by Jalpesh P. Vadgama
    In today’s computing world the world is all about Parallel processing. You have multicore CPU where you have different core doing different work parallel or its doing same task parallel. For example I am having 4-core CPU as follows. So the code that I write should take care of this.C# does provide that kind of facility to write code for multi core CPU with task parallel library. We will explore that in this post. Read More

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  • b2b SOA Suite partner training November 13th & 14th Bucharest

    - by JuergenKress
    Description: Oracle SOA Suite 11g is a complete infrastructure for building, deploying, and managing composite applications and business processes. For an enterprise to extend business processes to its trading partners, it requires a platform that addresses compliance, security, visibility, scalability, and standards. The Oracle SOA Suite (Oracle B2B) is this platform. Oracle B2B, the "Edge Component", enables an enterprise to define, configure, manage, and monitor the exchange of information, with its trading partners. Oracle SOA Suite, the "B2B Infrastructure", enables business process orchestration, administration, monitoring, auditing, inter-enterprise connectivity, governance and security. Together they provide a complete end-to-end business process integration platform. Date Location Time Facilitator Register 13-14 November Oracle Room MtgRm15_6, Bucharest - Nusco Tower, Romania D1: 08:30 - 17:30 D2: 09:00 - 17:30 Krishnaprem Bhatia Please contact us directly Registration Please contact us directly - Please note that there are limited seats and confirmation will be on a first come, first served basis Travel Each delegate is responsible for his/her own travel arrangements. Please obtain approval from your manager first. Contact For logistic questions, please contact Nadja Vogl SOA & BPM Partner Community For regular information on Oracle SOA Suite become a member in the SOA & BPM Partner Community for registration please visit  www.oracle.com/goto/emea/soa (OPN account required) If you need support with your account please contact the Oracle Partner Business Center. Blog Twitter LinkedIn Mix Forum Technorati Tags: b2b,SOA Suite,training,eduction,b2b training,SOA Community,Oracle SOA,Oracle BPM,Community,OPN,Jürgen Kress

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  • Secure against c99 and similar shells

    - by Amit Sonnenschein
    I'm trying to secure my server as much as i can without limiting my options, so as a first step i've prevented dangerous functions with php disable_functions = "apache_child_terminate, apache_setenv, define_syslog_variables, escapeshellarg, escapeshellcmd, eval, exec, fp, fput, ftp_connect, ftp_exec, ftp_get, ftp_login, ftp_nb_fput, ftp_put, ftp_raw, ftp_rawlist, highlight_file, ini_alter, ini_get_all, ini_restore, inject_code, mysql_pconnect, openlog, passthru, php_uname, phpAds_remoteInfo, phpAds_XmlRpc, phpAds_xmlrpcDecode, phpAds_xmlrpcEncode, popen, posix_getpwuid, posix_kill, posix_mkfifo, posix_setpgid, posix_setsid, posix_setuid, posix_setuid, posix_uname, proc_close, proc_get_status, proc_nice, proc_open, proc_terminate, shell_exec, syslog, system, xmlrpc_entity_decode" but i'm still fighting directory travel, i can't seems to be able to limit it, by using a shell script like c99 i can travel from my /home/dir to anywhere on the disc. how can i limit it once and for all ?

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  • Chiro One's Success with Oracle Sales Cloud

    - by Richard Lefebvre
    "As we strive to be a world-class company for our patients we needed to align with a company like Oracle who also is a world-class brand," says Roger Gold, National Director of Customer Engagement. Chiro One Wellness Centers is a national healthcare provider. It plans to grow to 1000 offices across the globe by 2020 and Oracle Sales Cloud will make that possible. Watch the 4'33 YouTube video featuring Chiro One success with Oracle Sales Cloud

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  • Questions to Know the Real SEO Experts

    Since the dawn of the internet, companies have eventually been making the shift from the physical world over to the virtual world, and being able to rank in a search engine is now the main factor that will determine if your business is a success. In the internet marketing universe, you probably are wondering how to know the real marketing experts from fake ones.

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  • Problem Implementing Texture on Libgdx Mesh of Randomized Terrain

    - by BrotherJack
    I'm having problems understanding how to apply a texture to a non-rectangular object. The following code creates textures such as this: from the debug renderer I think I've got the physical shape of the "earth" correct. However, I don't know how to apply a texture to it. I have a 50x50 pixel image (in the environment constructor as "dirt.png"), that I want to apply to the hills. I have a vague idea that this seems to involve the mesh class and possibly a ShapeRenderer, but the little i'm finding online is just confusing me. Bellow is code from the class that makes and regulates the terrain and the code in a separate file that is supposed to render it (but crashes on the mesh.render() call). Any pointers would be appreciated. public class Environment extends Actor{ Pixmap sky; public Texture groundTexture; Texture skyTexture; double tankypos; //TODO delete, temp public Tank etank; //TODO delete, temp int destructionRes; // how wide is a static pixel private final float viewWidth; private final float viewHeight; private ChainShape terrain; public Texture dirtTexture; private World world; public Mesh terrainMesh; private static final String LOG = Environment.class.getSimpleName(); // Constructor public Environment(Tank tank, FileHandle sfileHandle, float w, float h, int destructionRes) { world = new World(new Vector2(0, -10), true); this.destructionRes = destructionRes; sky = new Pixmap(sfileHandle); viewWidth = w; viewHeight = h; skyTexture = new Texture(sky); terrain = new ChainShape(); genTerrain((int)w, (int)h, 6); Texture tankSprite = new Texture(Gdx.files.internal("TankSpriteBase.png")); Texture turretSprite = new Texture(Gdx.files.internal("TankSpriteTurret.png")); tank = new Tank(0, true, tankSprite, turretSprite); Rectangle tankrect = new Rectangle(300, (int)tankypos, 44, 45); tank.setRect(tankrect); BodyDef terrainDef = new BodyDef(); terrainDef.type = BodyType.StaticBody; terrainDef.position.set(0, 0); Body terrainBody = world.createBody(terrainDef); FixtureDef fixtureDef = new FixtureDef(); fixtureDef.shape = terrain; terrainBody.createFixture(fixtureDef); BodyDef tankDef = new BodyDef(); Rectangle rect = tank.getRect(); tankDef.type = BodyType.DynamicBody; tankDef.position.set(0,0); tankDef.position.x = rect.x; tankDef.position.y = rect.y; Body tankBody = world.createBody(tankDef); FixtureDef tankFixture = new FixtureDef(); PolygonShape shape = new PolygonShape(); shape.setAsBox(rect.width*WORLD_TO_BOX, rect.height*WORLD_TO_BOX); fixtureDef.shape = shape; dirtTexture = new Texture(Gdx.files.internal("dirt.png")); etank = tank; } private void genTerrain(int w, int h, int hillnessFactor){ int width = w; int height = h; Random rand = new Random(); //min and max bracket the freq's of the sin/cos series //The higher the max the hillier the environment int min = 1; //allocating horizon for screen width Vector2[] horizon = new Vector2[width+2]; horizon[0] = new Vector2(0,0); double[] skyline = new double[width]; //TODO skyline necessary as an array? //ratio of amplitude of screen height to landscape variation double r = (int) 2.0/5.0; //number of terms to be used in sine/cosine series int n = 4; int[] f = new int[n*2]; //calculating omegas for sine series for(int i = 0; i < n*2 ; i ++){ f[i] = rand.nextInt(hillnessFactor - min + 1) + min; } //amp is the amplitude of the series int amp = (int) (r*height); double lastPoint = 0.0; for(int i = 0 ; i < width; i ++){ skyline[i] = 0; for(int j = 0; j < n; j++){ skyline[i] += ( Math.sin( (f[j]*Math.PI*i/height) ) + Math.cos(f[j+n]*Math.PI*i/height) ); } skyline[i] *= amp/(n*2); skyline[i] += (height/2); skyline[i] = (int)skyline[i]; //TODO Possible un-necessary float to int to float conversions tankypos = skyline[i]; horizon[i+1] = new Vector2((float)i, (float)skyline[i]); if(i == width) lastPoint = skyline[i]; } horizon[width+1] = new Vector2(800, (float)lastPoint); terrain.createChain(horizon); terrain.createLoop(horizon); //I have no idea if the following does anything useful :( terrainMesh = new Mesh(true, (width+2)*2, (width+2)*2, new VertexAttribute(Usage.Position, (width+2)*2, "a_position")); float[] vertices = new float[(width+2)*2]; short[] indices = new short[(width+2)*2]; for(int i=0; i < (width+2); i+=2){ vertices[i] = horizon[i].x; vertices[i+1] = horizon[i].y; indices[i] = (short)i; indices[i+1] = (short)(i+1); } terrainMesh.setVertices(vertices); terrainMesh.setIndices(indices); } Here is the code that is (supposed to) render the terrain. @Override public void render(float delta) { Gdx.gl.glClearColor(1, 1, 1, 1); Gdx.gl.glClear(GL10.GL_COLOR_BUFFER_BIT); // tell the camera to update its matrices. camera.update(); // tell the SpriteBatch to render in the // coordinate system specified by the camera. backgroundStage.draw(); backgroundStage.act(delta); uistage.draw(); uistage.act(delta); batch.begin(); debugRenderer.render(this.ground.getWorld(), camera.combined); batch.end(); //Gdx.graphics.getGL10().glEnable(GL10.GL_TEXTURE_2D); ground.dirtTexture.bind(); ground.terrainMesh.render(GL10.GL_TRIANGLE_FAN); //I'm particularly lost on this ground.step(); }

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  • how to calculate intersection time and place of multiple moving arcs

    - by user20733
    I have rocks orbiting moons, moons orbiting planets, planets orbiting suns, and suns orbiting black holes, and the current system could have many many layers of orbitage. the position of any object is a function of time and relative to the object it orbits. (so far so good). now I want to know for a given 2 objects(A,B), a start time and a speed, how can I work out the when and where to go. I can work out where A and B is given a time. so i just need. 1: direction to travel in from A to B(remember B is moving(not in a straight line)) 2: Time to get to b in a straight line. travel must be in a straight line with the shortest possible distance. as an extension to this question, how will i know if its better to wait, EG is it faster to stay on object A and wait for a hour when the objects may be closer, than to set off from A to B at the start. Cheers, it hurt my brain.

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  • SEO Gives You Free Traffic

    The World Wide Web can be such a daunting place to a newbie. Perhaps you have a website and you are wondering what all the fuss is about on the World Wide Web? Perhaps you are looking to start a new venture and think the internet could be a perfect place for it? Well how do you get your website visible to your target audience?

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  • Site Owners Must Hire a SEO Company For Their Websites

    The number of websites in the worldwide web is increasing rapidly, currently there are over billions of websites (some active and some inactive) in the internet world and it can be easily presumed that the number of website will increase to a large extent in future. Most of these websites have been published by companies and organizations in order to spread their business worldwide. As we all know, internet is the best and easiest way to spread any business all over the world.

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  • The Job of SEO Spiders

    The World Wide Web, also known as the Internet, is a very complex world. Search engines like Google need a software program that can read what's on the web. The said software program is known as bot or spider or crawler.

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  • TrueCrypt with blank password

    - by TheSavo
    This might sound like an odd question, but please hear me out. I want to setup my laptop to use TrueCrypt, but without a password. My wife doesn't travel much with her notebook. There are occasions where we will have to travel and bring the notebook with us. This is when we would like the have the encryption. The time it takes to to change the unlock pass-phrase is considerably less than it takes to encrypt the drive. My thoughts are to leave the disk encrypted but with no password. Then when it's time to pack for a trip and bring the laptop, we would then change the pass-phrase and create a rescue disk.

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  • Top 5 SEO Experts Tips For 2010

    The world of internet marketing and online advertising is so dynamic that it may change almost everyday. The change is not just limited to the content but to how the whole system operates. The rules, guidelines or tips that help you in cutting the competition and bringing out the best content to the world in the best possible manner can also change very rapidly.

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  • How to Opt For C Class IP Addresses

    There are a great number of SEO Hosting Services in the world today that are formulated with the intention of being able to create and develop winning strategies for their websites to rank well on the top search engines of the world. The need for SEO Hosting has come at a time when webmasters are seeking ways in which they can assign multiple C Class IP Addresses to all of their domains and mange as well as create some of the best services that are able to know where webmasters can control all their domains from one single cPanel.

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  • What is the Need of Website Development?

    In the corporate world, the danger of not having a strong web presence cannot be imagined by the businesses. Every business needs to be up to date, to be successful and boost up their business; it needs to have a place in the World Wide Web.

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