Ios 目标C-如何知道点是否在圆的四分之一以内?
我在一个矩形内画了四分之一个圆 矩形:Ios 目标C-如何知道点是否在圆的四分之一以内?,ios,objective-c,geometry,uibezierpath,Ios,Objective C,Geometry,Uibezierpath,我在一个矩形内画了四分之一个圆 矩形: UIView *Rectangle = [[UIView alloc] initWithFrame:CGRectMake(0,0,[[UIScreen mainScreen] bounds].size.width,[[UIScreen mainScreen] bounds].size.height-292)]; Rectangle.backgroundColor = [UIColor lightGrayColor]; Rectangle.
UIView *Rectangle = [[UIView alloc] initWithFrame:CGRectMake(0,0,[[UIScreen mainScreen] bounds].size.width,[[UIScreen mainScreen] bounds].size.height-292)];
Rectangle.backgroundColor = [UIColor lightGrayColor];
Rectangle.layer.zPosition = -5;
CGPoint center;
center.x = 0;
center.y = 0;
float radius = [[UIScreen mainScreen] bounds].size.width;
UIBezierPath *circle = [UIBezierPath bezierPathWithArcCenter:center
radius:radius
startAngle:0
endAngle:M_PI
clockwise:YES];
CAShapeLayer *circleLayer = [CAShapeLayer layer];
[circleLayer setPath:[circle CGPath]];
Cirlce的四分之一:
UIView *Rectangle = [[UIView alloc] initWithFrame:CGRectMake(0,0,[[UIScreen mainScreen] bounds].size.width,[[UIScreen mainScreen] bounds].size.height-292)];
Rectangle.backgroundColor = [UIColor lightGrayColor];
Rectangle.layer.zPosition = -5;
CGPoint center;
center.x = 0;
center.y = 0;
float radius = [[UIScreen mainScreen] bounds].size.width;
UIBezierPath *circle = [UIBezierPath bezierPathWithArcCenter:center
radius:radius
startAngle:0
endAngle:M_PI
clockwise:YES];
CAShapeLayer *circleLayer = [CAShapeLayer layer];
[circleLayer setPath:[circle CGPath]];
然后我在视图中添加了矩形,并在矩形内添加了圆形:
[self.view addSubview:Rectangle];
[Rectangle.layer addSublayer:circleLayer];
然后,我开始画1个宽度和1个高度的小矩形,我把它们看作是点,然后用一个for循环将它们随机地添加到视图中,用绿色表示圆圈内的点,并用红色
圆圈外的点。int compteurPointsinCercle = 0 ;
int compteurPointsOutCercle = 0 ;
float XcenterCircle = center.x;
float YcenterCircle = center.y;
for (int i = 0 ; i < 50000 ; i++ )
{
float xvalue = arc4random_uniform([[UIScreen mainScreen] bounds].size.width);
float yvalue = arc4random_uniform([[UIScreen mainScreen] bounds].size.height-292);
// (x - center_x)^2 + (y - center_y)^2 < radius^2
float valeurPoint = (xvalue - XcenterCircle)*2 + (yvalue -YcenterCircle)*2;
NSLog(@"(Inside for), valeurPoint is : %f",valeurPoint);
if ( valeurPoint < (radius*2) )
{
// Point is inside of circle (green color)
compteurPointsinCercle++;
UIView *Rectangle2 = [[UIView alloc] initWithFrame:CGRectMake(xvalue,yvalue,1,1)];
Rectangle2.backgroundColor = [UIColor greenColor];
[self.view addSubview:Rectangle2];
}
else if ( valeurPoint > (radius*2) )
{
// Point is outside of circle (red color)
compteurPointsOutCercle++;
UIView *Rectangle2 = [[UIView alloc] initWithFrame:CGRectMake(xvalue,yvalue,1,1)];
Rectangle2.backgroundColor = [UIColor redColor];
[self.view addSubview:Rectangle2];
}
}
其中,xvalue
和yvalue
是将要创建的点的坐标,XcenterCircle
和YcenterCircle
是圆心的坐标
我有点不对劲,因为它给出了这个结果(如果该点在圆内或不在圆内,则测试不正确:圆内的一部分点被视为在圆外):
你能告诉我我做错了什么吗?我怎样才能精确地计算出圆内的点呢
*
不是幂运算,而是乘法
float valeurPoint = (xvalue - XcenterCircle) * (xvalue - XcenterCircle) + (yvalue -YcenterCircle)*(yvalue -YcenterCircle);
if ( valeurPoint < (radius * radius) )
您也可以直接使用hypot
函数(尽管由于sqrt
计算,性能可能稍差)
float distance=hypotf((xvalue-XcenterCircle),(yvalue-YcenterCircle));
if(距离<半径)
编辑:
谢谢@Alex的建议。最好的解决方案是使用本机方法
-[uiberpath containsPoint:][/code>。那你就不必计算距离了。可能是这样的:@Alex这是一个很好的解决方案。我将测试这个解决方案,然后回复你。我是为我做的,我在电源操作方面遇到了问题
float distance = hypotf((xvalue - XcenterCircle), (yvalue -YcenterCircle));
if (distance < radius)