3D立方体问题！第1部分 [英] 3D Cube Problem! Part 1

问题描述

我已经使用 CALayer的在iPhone中创建了3D立方体。现在我想在用户双击时以90˚的角度旋转该多维数据集（ CALayer ）。

I have created a 3D cube in iphone using CALayer's. Now I wanted to rotate that cube(CALayer) at 90˚ when user double taps on it.

我能够将那个多维数据集（ CALayer ）旋转到90˚一次，但是当我双击多维数据集（ CALayer ）没有旋转时。

I was able to rotate that cube(CALayer) to 90˚ once but when I double tap the cube(CALayer) is not rotating.

这是我使用的代码旋转立方体（ CALayer ）

Here is the code that I used to rotate the cube(CALayer)

CATransform3D x = CATransform3DRotate(currentLayer.sublayerTransform, M_PI / 2, 0, 0, 1);
currentLayer.transform = x;

任何人都可以帮忙。我做错了。

Can anyone help in this. What I'm doing wrong.

PS。对于想知道我如何获得学位的人来说，这就是窍门

PS. For the people who are wondering how I got the degree sign then here is the trick

选项 + K

推荐答案

它是因为您没有更改旋转角度....要理解这一点，可以说您正在传递每个M_PI / 2该方法的时间....因此， CATransform3DRotate 不要将其旋转到 next90˚，而是将图层旋转到指定的角度在这种情况下为90 ...因此您不会遇到任何障碍，因为它已经在90˚ ......为了获得正确的结果，请执行此操作

its because you are not changing the angle of rotation .... to understand this lets say you are passing M_PI/2 each time to that method .... so CATransform3DRotate do not rotate it to next 90˚ rather it rotate the layer to the specified angle in this case its 90... so you are not getting any chage because it already at 90˚ ..... so to get correct result do this

static float angle = M_PI / 2;//dont make it static rather make it a global variable
angle += M_PI / 2;
CATransform3D x = CATransform3DRotate(currentLayer.sublayerTransform,angle, 0, 0, 1);
currentLayer.transform = x;

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