稍有opengl或3d基础的都知道平移/旋转/缩放这几个基本模型视图变换的实现原理, 最近看了下cocos2d-x相关部分的实现, 了解了这些实现那些各种坐标变换基本不在话下了, cocos2d-x本身还是相对简单的引擎.
1. CCAffineTransform
struct CCAffineTransform {
float a, b, c, d;
float tx, ty;
};
表示变换矩阵:
构造CCAffineTransform结构
CCAffineTransform __CCAffineTransformMake(float a, float b, float c, float d, float tx, float ty)
{
CCAffineTransform t;
t.a = a; t.b = b; t.c = c; t.d = d; t.tx = tx; t.ty = ty;
return t;
}
2. 单位矩阵
CCAffineTransform CCAffineTransformMakeIdentity()
{
return __CCAffineTransformMake(1.0, 0.0, 0.0, 1.0, 0.0, 0.0);
}
将CCAffineTransform构造为单位矩阵:
3. 平移
CCAffineTransform CCAffineTransformTranslate(const CCAffineTransform& t, float tx, float ty)
{
return __CCAffineTransformMake(t.a, t.b, t.c, t.d, t.tx + t.a * tx + t.c * ty, t.ty + t.b * tx + t.d * ty);
}
将CCAffineTransform矩阵和平移矩阵的结果：
4. 旋转
CCAffineTransform CCAffineTransformRotate(const CCAffineTransform& t, float anAngle)
{
float fSin = sin(anAngle);
float fCos = cos(anAngle); return __CCAffineTransformMake( t.a * fCos + t.c * fSin,
t.b * fCos + t.d * fSin,
t.c * fCos - t.a * fSin,
t.d * fCos - t.b * fSin,
t.tx,
t.ty);
}
绕Z轴旋转矩阵右乘以变换矩阵：
5. 缩放
CCAffineTransform CCAffineTransformScale(const CCAffineTransform& t, float sx, float sy)
{
return __CCAffineTransformMake(t.a * sx, t.b * sx, t.c * sy, t.d * sy, t.tx, t.ty);
}
6. Concate
/* Concatenate `t2' to `t1' and return the result:
t' = t1 * t2 */
CCAffineTransform CCAffineTransformConcat(const CCAffineTransform& t1, const CCAffineTransform& t2)
{
return __CCAffineTransformMake( t1.a * t2.a + t1.b * t2.c, t1.a * t2.b + t1.b * t2.d, //a,b
t1.c * t2.a + t1.d * t2.c, t1.c * t2.b + t1.d * t2.d, //c,d
t1.tx * t2.a + t1.ty * t2.c + t2.tx, //tx
t1.tx * t2.b + t1.ty * t2.d + t2.ty); //ty
}
结果相当于t2 . t1
7. CCPointApplyAffineTransform
CCPoint __CCPointApplyAffineTransform(const CCPoint& point, const CCAffineTransform& t)
{
CCPoint p;
p.x = (float)((double)t.a * point.x + (double)t.c * point.y + t.tx);
p.y = (float)((double)t.b * point.x + (double)t.d * point.y + t.ty);
return p;
}
8. CCAffineTransformInvert
CCAffineTransform CCAffineTransformInvert(const CCAffineTransform& t)
{
float determinant = / (t.a * t.d - t.b * t.c); return __CCAffineTransformMake(determinant * t.d, -determinant * t.b, -determinant * t.c, determinant * t.a,
determinant * (t.c * t.ty - t.d * t.tx), determinant * (t.b * t.tx - t.a * t.ty) );
}
求矩阵的逆矩阵，通过Mathematica计算得：