仅对世界上的一个精灵施加零重力

Question

我在 cocos2d 和 box2d 中有一个世界，它有重力。 现在，为了为每个精灵添加主体，我调用一个函数并向她发送精灵。

sprite1，必须根据重力移动，但 sprite2 必须是静态的，没有重力，直到 sprite1 击中它，然后世界力量才会影响他。

我如何仅将精灵1/身体重力设置为零，直到另一个精灵击中他？

我的问题是所有精灵对身体使用相同的功能：

- (void)addBoxBodyForSprite:(CCSprite *)sprite {

    b2BodyDef spriteBodyDef;
    spriteBodyDef.type = b2_dynamicBody;
    spriteBodyDef.position.Set(sprite.position.x/PTM_RATIO,sprite.position.y/PTM_RATIO);
    spriteBodyDef.userData = sprite;
    spriteBody = world->CreateBody(&spriteBodyDef);

    b2PolygonShape spriteShape;
    spriteShape.SetAsBox(sprite.contentSize.width/PTM_RATIO/2,sprite.contentSize.height/PTM_RATIO/2);
    b2FixtureDef spriteShapeDef;
    spriteShapeDef.shape = &spriteShape;
    spriteShapeDef.density = 10.0;
    spriteShapeDef.isSensor = true;
    spriteBody->CreateFixture(&spriteShapeDef);

}

我想在开始时仅在精灵1上应用重力，但我确实想为精灵2创建一个身体，因为稍后它将受到世界的影响。

那么，在开始时创建 2 个物体后，如何仅阻止 sprite2 掉落？
多谢。

Answer 1

我通常最常使用 SetActive()，但根据您的需求，我认为 SetAwake() 就是您想要的。

b2Body.h

/// You can disable sleeping on this body. If you disable sleeping, the
/// body will be woken.
void SetSleepingAllowed(bool flag);

/// Is this body allowed to sleep
bool IsSleepingAllowed() const;

/// Set the sleep state of the body. A sleeping body has very
/// low CPU cost.
/// @param flag set to true to put body to sleep, false to wake it.
void SetAwake(bool flag);

/// Get the sleeping state of this body.
/// @return true if the body is sleeping.
bool IsAwake() const;

/// Set the active state of the body. An inactive body is not
/// simulated and cannot be collided with or woken up.
/// If you pass a flag of true, all fixtures will be added to the
/// broad-phase.
/// If you pass a flag of false, all fixtures will be removed from
/// the broad-phase and all contacts will be destroyed.
/// Fixtures and joints are otherwise unaffected. You may continue
/// to create/destroy fixtures and joints on inactive bodies.
/// Fixtures on an inactive body are implicitly inactive and will
/// not participate in collisions, ray-casts, or queries.
/// Joints connected to an inactive body are implicitly inactive.
/// An inactive body is still owned by a b2World object and remains
/// in the body list.
void SetActive(bool flag);

/// Get the active state of the body.
bool IsActive() const;

这应该就是您需要的一切。

Answer 2

您可以尝试仅将您想要受世界影响的精灵添加到空间中，然后再添加您想要受世界影响的精灵。

因此，在这种情况下，将 Sprite1 添加到您的空间，但不添加 Sprite2，稍后您可以添加 Sprite2，这样它就会受到空间的影响。