我想平滑我的手繪製貝塞爾曲線,但無法實現,它,我從我試圖實現的埃裏卡蘇丹寫的書中得到了平滑曲線的代碼,但是不知道爲什麼我不能夠產生平滑的曲線,下面是我的代碼,我要求所有的你,請通過它..平滑手繪貝塞爾曲線
- (void)drawRect:(CGRect)rect
{
myPath.pathColor = [UIColor redColor];
for (BezeirPath *_path in m_pathArray)
{
[_path.pathColor setStroke];
[_path.bezeirPath strokeWithBlendMode:kCGBlendModeNormal alpha:1.0];
}
}
- (void)touchesMoved:(NSSet *)touches withEvent:(UIEvent *)event
{
UITouch *mytouch=[[touches allObjects] objectAtIndex:0];
m_previousPoint1 = [mytouch previousLocationInView:self];
m_previousPoint2 = [mytouch previousLocationInView:self];
m_currentPoint = [mytouch locationInView:self];
CGPoint controlPoint = CGPointMake(m_previousPoint1.x+(m_previousPoint1.x - m_previousPoint2.x), m_previousPoint1.y +(m_previousPoint1.y - m_previousPoint2.y));
[myPath.bezeirPath addQuadCurveToPoint:m_currentPoint controlPoint:controlPoint];
self.previousPoint3 = controlPoint;
[myPath.bezeirPath smoothedPath:myPath.bezeirPath :40];//Called the smoothing function.
[self setNeedsDisplay];
}
//平滑功能
-(UIBezierPath*)smoothedPath:(UIBezierPath*)bpath: (NSInteger) granularity
{
NSArray *points = pointsFromBezierPath(bpath);
if (points.count < 4) return bpath;
// This only copies lineWidth. You may want to copy more
UIBezierPath *smoothedPath = [UIBezierPath bezierPath];
smoothedPath.lineWidth = bpath.lineWidth;
// Draw out the first 3 points (0..2)
[smoothedPath moveToPoint:POINT(0)];
for (int index = 1; index < 3; index++)
[smoothedPath addLineToPoint:POINT(index)];
// Points are interpolated between p1 and p2,
// starting with 2..3, and moving from there
for (int index = 4; index < points.count; index++)
{
CGPoint p0 = POINT(index - 3);
CGPoint p1 = POINT(index - 2);
CGPoint p2 = POINT(index - 1);
CGPoint p3 = POINT(index);
// now add n points starting at p1 + dx/dy up until p2
// using Catmull-Rom splines
for (int i = 1; i < granularity; i++)
{
float t = (float) i * (1.0f/(float) granularity);
float tt = t * t;
float ttt = tt * t;
CGPoint pi; // intermediate point
pi.x = 0.5 * (2*p1.x+(p2.x-p0.x)*t +
(2*p0.x-5*p1.x+4*p2.x-p3.x)*tt +
(3*p1.x-p0.x-3*p2.x+p3.x)*ttt);
pi.y = 0.5 * (2*p1.y+(p2.y-p0.y)*t +
(2*p0.y-5*p1.y+4*p2.y-p3.y)*tt +
(3*p1.y-p0.y-3*p2.y+p3.y)*ttt);
[smoothedPath addLineToPoint:pi];
}
[smoothedPath addLineToPoint:p2];
}
// finish by adding the last point
[smoothedPath addLineToPoint:POINT(points.count - 1)];
return smoothedPath;
}
不,我必須將bezeirPath的對象傳遞給此函數。糾正我,如果我錯了。 – Ranjit
你好朋友,可以有人建議我在這方面 – Ranjit