如何获得旋转的CCNode的角坐标？

Question

我有一个CCNode与一定的旋转。为了简单起见，可以说它旋转了45度，尺寸是100 * 100，位置是200,200。

我期待找到4个坐标（标记为绿色）在该图像上。在框架中是否有任何方法可以用于此操作，还是需要用三角函数手动执行？在那种情况下，我该怎么做最简单的方法？

的最好的事情是，如果我能做出这样的CCNode类，所以它的方便的方法。

Answer 1

CCNode.h：

/** Converts a Point to world space coordinates. The result is in Points. 
@since v0.7.1 
*/ 
- (CGPoint)convertToWorldSpace:(CGPoint)nodePoint; 

/** Converts a Point to node (local) space coordinates. The result is in Points. 
@since v0.7.1 
*/ 
- (CGPoint)convertToNodeSpace:(CGPoint)worldPoint; 

例子：

// top right corner of your node 
CGPoint topRight = ccp(node.contentSize.width, node.contentSize.height); 
// same point in world coordinates 
CGPoint topRightWorld = [node convertToWorldSpace:topRight]; 

Answer 2

试试这个：

CGFloat angle = node.rotation * M_PI/180; 
CGRect frame = CGRectMake(node.position.x, node.position.y, node.contentSize.width, node.contentSize.height); 

CGPoint ip0 = CGPointMake(frame.origin.x,frame.origin.y); 
CGPoint ip1 = CGPointMake(frame.origin.x +frame.size.width, 0); 
CGPoint ip2 = CGPointMake(0,frame.origin.y +frame.size.height); 
CGPoint ip3 = CGPointMake(frame.origin.x +frame.size.width,frame.origin.y +frame.size.height); 

ip0 = CGPointApplyAffineTransform(ip0, CGAffineTransformMakeRotation(angle)); 
ip1 = CGPointApplyAffineTransform(ip1, CGAffineTransformMakeRotation(angle)); 
ip2 = CGPointApplyAffineTransform(ip2, CGAffineTransformMakeRotation(angle)); 
ip3 = CGPointApplyAffineTransform(ip3, CGAffineTransformMakeRotation(angle));