限制CGAffineTransform规模

Question

我正在使用CGAffineTransformMake转换视图.它旋转,缩放和翻译.这很好用.但我无法找到一种方法将规模限制到最大尺寸.

如果超出比例,我仍然需要应用当前的旋转和平移.

任何建议都非常感谢!

资源:

UITouch *touch1 = [sortedTouches objectAtIndex:0];
UITouch *touch2 = [sortedTouches objectAtIndex:1];

CGPoint beginPoint1 = *(CGPoint *)CFDictionaryGetValue(touchBeginPoints, touch1);
CGPoint currentPoint1 = [touch1 locationInView:self.superview];
CGPoint beginPoint2 = *(CGPoint *)CFDictionaryGetValue(touchBeginPoints, touch2);
CGPoint currentPoint2 = [touch2 locationInView:self.superview];

double layerX = self.center.x;
double layerY = self.center.y;

double x1 = beginPoint1.x - layerX;
double y1 = beginPoint1.y - layerY;
double x2 = beginPoint2.x - layerX;
double y2 = beginPoint2.y - layerY;
double x3 = currentPoint1.x - layerX;
double y3 = currentPoint1.y - layerY;
double x4 = currentPoint2.x - layerX;
double y4 = currentPoint2.y - layerY;

// Solve the system:
//   [a b t1, -b a t2, 0 0 1] * [x1, y1, 1] = [x3, y3, 1]
//   [a b t1, -b a t2, 0 0 1] * [x2, y2, 1] = [x4, y4, 1]

double D = (y1-y2)*(y1-y2) + (x1-x2)*(x1-x2);
if (D < 0.1) {
    return CGAffineTransformMakeTranslation(x3-x1, y3-y1);
}

double a = (y1-y2)*(y3-y4) + (x1-x2)*(x3-x4);
double b = (y1-y2)*(x3-x4) - (x1-x2)*(y3-y4);
double tx = (y1*x2 - x1*y2)*(y4-y3) - (x1*x2 + y1*y2)*(x3+x4) + x3*(y2*y2 + x2*x2) + x4*(y1*y1 + x1*x1);
double ty = (x1*x2 + y1*y2)*(-y4-y3) + (y1*x2 - x1*y2)*(x3-x4) + y3*(y2*y2 + x2*x2) + y4*(y1*y1 + x1*x1);

return   CGAffineTransformMake(a/D, -b/D, b/D, a/D, tx/D, ty/D);

Answer 1

做这样的事情:

CGAffineTransform transform = self.view.transform;
float scale = sqrt(transform.a*transform.a + transform.c*transform.c);
if (scale > SCALE_MAX)
    self.view.transform = CGAffineTransformScale(transform, SCALE_MAX/scale, SCALE_MAX/scale);
else if (scale < SCALE_MIN)
    self.view.transform = CGAffineTransformScale(transform, SCALE_MIN/scale, SCALE_MIN/scale);

在touchesMoved:withEvent:和updateOriginalTransformForTouches方法结束时.基本上,您检查当前比例是否超过某个SCALE_MAX值,然后将转换矩阵与反转比例值相乘.