一名工会B再减去一个交点B

Question

我看到有提供矩形

CGRect CGRectIntersection(CGRect r1, CGRect r2) 
CGRectUnion(CGRect r1, CGRect r2) 

方法，但在那里，提供有一个工会B减一路口B.凡区域或坐标列表的方法A是一个大的矩形，B是一个完整的小矩形？

Answer 1

您将如何表示坐标列表，您将使用哪种数据类型？ CGRects使用浮点坐标，所以显然你不能有一组（x，y）点。我想你会把联合减去交叉点分解成一堆单独的CGRects，但它看起来很麻烦。也许你可以更清楚地知道你想要达到的目标？

要尝试回答你的问题，怎么样是这样的：

BOOL CGPointIsInUnionButNotIntersectionOfRects(CGRect r1, CGRect r2, CGPoint p) 
{ 
    CGRect union = CGRectUnion(r1, r2); 
    CGRect intersection = CGRectIntersection(r1, r2); 
    return CGRectContainsPoint(union) && !CGRectContainsPoint(instersection); 
} 

Answer 2

我猜你的意思是“提供区域”，“画到屏幕”：使用CGPath有两个矩形和奇偶填充规则。

// create a graphics context with the C-API of the QuartzCore framework 
// the graphics context is only required for drawing the subsequent drawing 
CGRect compositionBounds = CGRectMake(30, 30, 300, 508); 
UIGraphicsBeginImageContext(compositionBounds.size); 
CGContextRef ctx = UIGraphicsGetCurrentContext(); 

// create a mutable path with QuartzCore 
CGMutablePathRef p1 = CGPathCreateMutable(); 
CGRect r1 = CGRectMake(20, 300, 200, 100); 
CGPathAddRect(p1, nil, r1); 
CGPathAddRect(p1, nil, CGRectInset(r1, 10, 10)); 

// draw our mutable path to the context filling its area 
CGContextAddPath(ctx, p1); 
CGContextSetFillColorWithColor(ctx, [[UIColor redColor] CGColor]); 
CGContextEOFillPath(ctx); 

// display our mutable path in an image view on screen 
UIImage *i = UIGraphicsGetImageFromCurrentImageContext(); 
UIImageView *iv = [[UIImageView alloc] initWithImage:i]; 
[self.view addSubview:iv]; 
iv.frame = self.view.frame; 

// release ressources created with the C-API of QuartzCore 
CGPathRelease(p1); 
UIGraphicsEndImageContext();