Question

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我试图通过libGDX在Box2D中产生一个Pong克隆。 Null Pointer Exception源自ContactListener的beginContact（）方法，我在其中尝试实现得分逻辑。

Box2D世界使用两个独立的EdgeShapes作为传感器，用于在与Ball碰撞时递增得分变量（查看附加图像）。 Ball和两个EdgeShapes之间的碰撞逻辑有效，但Ball和Box2D世界中任何其他东西之间的碰撞都会使程序崩溃。

enter image description here

堆栈跟踪：

Exception in thread "LWJGL Application" java.lang.NullPointerException
at com.ckq3r.Pong.screens.GameScreen$2.beginContact(GameScreen.java:491)
at com.badlogic.gdx.physics.box2d.World.beginContact(World.java:876)
at com.badlogic.gdx.physics.box2d.World.jniStep(Native Method)
at com.badlogic.gdx.physics.box2d.World.step(World.java:602)
at com.ckq3r.Pong.screens.GameScreen.render(GameScreen.java:99)
at com.badlogic.gdx.Game.render(Game.java:46)
at com.ckq3r.Pong.PongGame.render(PongGame.java:236)
at com.badlogic.gdx.backends.lwjgl.LwjglApplication.mainLoop(LwjglApplication.java:204)
at com.badlogic.gdx.backends.lwjgl.LwjglApplication$1.run(LwjglApplication.java:112)

有问题的代码：

    /**Box2D contact listener**/
private void createContactListener() {
    world.setContactListener(new ContactListener() {

        @Override
        public void beginContact(Contact contact) {
            Fixture fixtureA = contact.getFixtureA();
            Fixture fixtureB = contact.getFixtureB();

            Gdx.app.log("beginContact", "between " + fixtureA.toString() + " and " + fixtureB.toString());

            if(fixtureA.getBody().getUserData().equals(1) && fixtureB.getBody().getUserData().equals(2) || fixtureA.getBody().getUserData().equals(2) && fixtureB.getBody().getUserData().equals(1)){
                Gdx.app.log("HIT", "goal1 contact");
                score1++;
                score1String = score1 + "";
            }

            if(fixtureA.getBody().getUserData().equals(1) && fixtureB.getBody().getUserData().equals(3) || fixtureA.getBody().getUserData().equals(3) && fixtureB.getBody().getUserData().equals(1)){
                Gdx.app.log("HIT", "goal2 contact");
                score2++;
                score2String = score2 + "";
            }
        }

        @Override
        public void endContact(Contact contact) {
            Fixture fixtureA = contact.getFixtureA();
            Fixture fixtureB = contact.getFixtureB();
            Gdx.app.log("endContact", "between " + fixtureA.toString() + " and " + fixtureB.toString());
        }

        @Override
        public void preSolve(Contact contact, Manifold oldManifold) {
        }

        @Override
        public void postSolve(Contact contact, ContactImpulse impulse) {
        }

        });
}

注意：

当我在beginContact（）方法中注释掉两个条件语句时，代码会运行。取消注释后，将再现错误。
Ball userData是circleBody.setUserData（1）;
goal1 userData是goalBody.setUserData（2）;
goal2 userData是goalBody.setUserData（3）;

Answer 1

在Box2D世界中为每个正文设置userdata，以便您可以在ContactListener中使用条件逻辑来评估是否应该处理碰撞。

重点放在下面的paddleBody.setUserData(4);行。您可以选择任何数字来设置单个对象的用户数据，只要该数字与您所在世界中其他对象的用户数据唯一。

例如：

    /**paddle1**/
    BodyDef paddleDef = new BodyDef();
    paddleDef.type = BodyType.DynamicBody;
    paddleDef.position.set(width * 0.2f, height / 2);
    paddleDef.fixedRotation = true;

    Body paddleBody = world.createBody(paddleDef);
    paddleBody.setUserData(4);

    PolygonShape paddleShape = new PolygonShape();
    paddleShape.setAsBox(3.0f, 15.0f); //half-width and half-height

    FixtureDef paddleFD = new FixtureDef();
    paddleFD.shape = paddleShape;
    paddleFD.density = 10.0f;
    paddleFD.friction = 0.0f;
    paddleFD.restitution = 0.1f;

    paddleBody.createFixture(paddleFD);

    paddleShape.dispose();

    paddleBody.setLinearVelocity(0.0f, 0.0f);
    /**end paddle1**/

    /**paddle2**/
    BodyDef paddleDef2 = new BodyDef();
    paddleDef2.type = BodyType.DynamicBody;
    paddleDef2.position.set(width * 0.8f, height / 2);
    paddleDef2.fixedRotation = true;        

    Body paddleBody2 = world.createBody(paddleDef2);
    paddleBody2.setUserData(5);             

    PolygonShape paddleShape2 = new PolygonShape();
    paddleShape2.setAsBox(3.0f, 15.0f); //half-width and half-height

    FixtureDef paddleFD2 = new FixtureDef();
    paddleFD2.shape = paddleShape2;
    paddleFD2.density = 10.0f;
    paddleFD2.friction = 0.0f;
    paddleFD2.restitution = 0.1f;

    paddleBody2.createFixture(paddleFD2);

    paddleShape2.dispose();

    paddleBody2.setLinearVelocity(0.0f, 0.0f); // Move nowhere at a rate of x = 0.0f, y = 0.0f meters per second
    /**end paddle2**/

如何处理libGDX的Box2D ContactListener中的空指针异常？

1 个答案: