CAShapeLayer的框架和界限 [英] Frame and Bounds of CAShapeLayer

问题描述

我正在处理 CAShapeLayer 。并尝试绘制非线性路径。我想将帧设置为 CAShapeLayer 。因此，我可以使用 CGPathGetPathBoundingBox 方法从 CGPathRef 获取帧。

I am working on CAShapeLayer.And trying to draw non-linear path.I want to set frame to CAShapeLayer.So i can use CGPathGetPathBoundingBox method to get frame from CGPathRef.

以下是代码：

CGMutablePathRef path = CGPathCreateMutable();
CGPathAddArc(path, NULL, rect.size.width/2, rect.size.height/2, 100, (0), (M_PI_2), NO);
CGPathAddArc(path, NULL, rect.size.width/2, rect.size.height/2, 100-50, (M_PI_2), (0), YES);
CGPathCloseSubpath(path);
CAShapeLayer* arcLayer = [[CAShapeLayer alloc]init];
arcLayer.path = path;
arcLayer.frame = CGPathGetPathBoundingBox(path);
arcLayer.bounds = CGPathGetPathBoundingBox(path);
[self.layer addSublayer:arcLayer]; ` 

请仔细参考我的代码。我已经设置了相同的框架和界限到CAShapeLayer。我的问题是如果我没有设置边界（与框架相同），然后它不会显示我的内容或它不会在框架内显示我的内容。为什么？请帮助我。谢谢你。

Please refer my code carefully .I have set same frame and bounds to CAShapeLayer.My problem is if i am not setting bounds (same as frame) ,then it wont show my content or it wont show my content within frame.Why?Please help me.Thanking you.

推荐答案

当您修改 CALayer 的框架时，您正在修改其尺寸并在其超级层的坐标空间中定位。 frame 是一个计算值，基于对象的界限和位置。在这种情况下，该层还没有超层，因此其坐标空间是无量纲的。设置框架时，大小会传递到边界属性，但由于无量纲空间中的任何点为零，位置将保持为零而不是您想要的位置。

When you modify the frame of a CALayer, you are modifying its size and position in the coordinate space of its superlayer. frame is a computed value, based on the object's bounds and position. In this case, the layer does not yet have a superlayer so its coordinate space is dimensionless. When you set the frame, the size gets carried through to the bounds property, but since any point in a dimensionless space is zero, the position will remain zero instead of what you'd intended.

要解决此问题，您可以设置边界在超级层的坐标空间中设置图层的大小和位置（这是一个真正的非计算属性），然后再将其添加为子图层。

To solve this, you can set the bounds to set the layer's size, and position (which is a real non-computed property) in the coordinate space of the superlayer before adding it as a sublayer.

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