我正在用 cocos2d、box2d 为 iPhone 开发一款 iPhone 游戏。我这里有一个问题。
从上面的截图可以看出,这些球是 b2Body。一切顺利,除了一件事。当我点击一个球时,我希望将其从屏幕上移除,并且效果很好。
但是,我也需要删除所有具有相同颜色的连接球。例如,当我点击底部的红球时,与这个被点击的球相交的每个红球也应该被移除。
此方法管理点击位置
- (void)ccTouchesEnded:(NSSet *)touches withEvent:(UIEvent *)event {
for( UITouch *touch in touches ) {
CGPoint location = [touch locationInView: [touch view]];
location = [[CCDirector sharedDirector] convertToGL: location];
b2Vec2 locationWorld = b2Vec2(location.x/PTM_RATIO, location.y/PTM_RATIO);
// go through every single b2Body
for(b2Body *b = world->GetBodyList(); b; b=b->GetNext()) {
if (b->GetUserData() != NULL) {
b2Fixture *bf = b->GetFixtureList();
// check which ball is tapped
if (bf->TestPoint(locationWorld)) {
[self destroyBall:b];
}
}
}
}
“destroyBall”方法是递归方法,写法如下
- (void) destroyBall:(b2Body *)body_ {
CCSprite *bodySprite = (CCSprite *) body_->GetUserData();
CGFloat h2 = bodySprite.contentSize.height / 2;
CGFloat w2 = bodySprite.contentSize.width / 2;
b2Vec2 p1 = b2Vec2((bodySprite.position.x - w2) / PTM_RATIO, (bodySprite.position.y - h2) / PTM_RATIO);
b2Vec2 p2 = b2Vec2((bodySprite.position.x - w2) / PTM_RATIO, (bodySprite.position.y + h2) / PTM_RATIO);
b2Vec2 p3 = b2Vec2((bodySprite.position.x + w2) / PTM_RATIO, (bodySprite.position.y + h2) / PTM_RATIO);
b2Vec2 p4 = b2Vec2((bodySprite.position.x + w2) / PTM_RATIO, (bodySprite.position.y - h2) / PTM_RATIO);
// go through all b2Body to check which one is intersect with body_
for (b2Body *b = world->GetBodyList(); b; b = b->GetNext()) {
if ((b->GetUserData() != NULL) && (b != body_)) {
CCSprite *newBall = (CCSprite *)b->GetUserData();
b2Fixture *bf = b->GetFixtureList();
if (bf->TestPoint(p1) || bf->TestPoint(p2) || bf->TestPoint(p3) || bf->TestPoint(p4)) {
if (bodySprite.tag == newBall.tag) {
[self destroyBall:b];
}
}
}
}
CCSpriteBatchNode *batch = (CCSpriteBatchNode *) [self getChildByTag:kTagBatchNode];
[batch removeChild:bodySprite cleanup:YES];
world->DestroyBody(body_);
}
这个想法很简单,当A球被轻敲时,它会寻找周围正在击球A球的其他球。如果是B球,则该过程将再次与B球一起进行,直到没有击球球为止。
但是,递归函数似乎不停地运行:(
如果您有任何想法或任何算法来使用 box2d 执行此功能,我将不胜感激
非常感谢您的支持。
唐