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        矩形和圆形碰撞检测

        Rectangle and Circle collision detection(矩形和圆形碰撞检测)

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                • 本文介绍了矩形和圆形碰撞检测的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

                  问题描述

                  我正在尝试在矩形和圆形之间进行碰撞检测.我想出了这个方法:

                  I am trying to do collision detection between a rectangle and a circle. I came up with this method:

                  -(BOOL) isCollidingRect:(CCSprite *) spriteOne WithSphere:(CCSprite *) spriteTwo {
                      float diff = ccpDistance(spriteOne.position, spriteTwo.position);
                      float obj1Radii = [spriteOne boundingBox].size.width/2;
                      float obj2Radii = [spriteTwo boundingBox].size.width/2;
                      if (diff < obj1Radii + obj2Radii) {
                          return YES;
                      } else {
                          return NO;
                      }
                  }
                  

                  这就是我检查它的方式:

                  and this is how I check it:

                  if ([self isCollidingRect:player WithSphere:blocker] == true) {
                     [self playerdeathstart];
                  }
                  

                  这似乎在矩形的一侧正常工作,但它不在它的上方或下方.在顶部和底部,碰撞发生得太早了.

                  This seems to work properly on the side of the rectangle but it doesn't above or below it. On the top and bottom, the collision occurs too early.

                  有没有办法可以正确检测到这种碰撞?感谢您的帮助.

                  Is there a way I can get this collision to detected properly? Thank you for your help.

                  推荐答案

                  你可以使用CGRectIntersectsRect 来实现这个.

                  You can use CGRectIntersectsRect to achieve this.

                  -(BOOL) isCollidingRect:(CCSprite *) spriteOne WithSphere:(CCSprite *) spriteTwo {
                      return CGRectIntersectsRect([spriteOne boundingBox],[spriteTwo boundingBox]);
                  }
                  

                  它不是像素完美的,但据我所知,在这种情况下没有必要.

                  It is not pixel perfect but as i understand that is not necessary in this case.

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