【发布时间】：2012-01-31 22:15:57
【问题描述】：
如何在 iOS 绘图应用程序 WHILE MOVING 中平滑一组点？我已经尝试过 UIBezierpaths，但是当我将点 1、2、3、4 - 2、3、4、5 移动时，我得到的只是它们相交的锯齿状末端。我听说过样条曲线和所有其他类型。我对 iPhone 编程很陌生，不明白如何在我的石英绘图应用程序中对其进行编程。一个可靠的例子将不胜感激，我已经花了几周的时间在圈子里跑，我似乎永远找不到任何 iOS 代码来完成这项任务。大多数帖子只是链接到一个 java 模拟或维基百科上关于曲线拟合的页面，这对我没有任何作用。我也不想切换到openGL ES。我希望最终有人可以提供代码来回答这个循环问题。
这是我的 UIBezierPath 代码，它在交叉路口留下边缘 ///
更新到下面的答案
#define VALUE(_INDEX_) [NSValue valueWithCGPoint:points[_INDEX_]]
#define POINT(_INDEX_) [(NSValue *)[points objectAtIndex:_INDEX_] CGPointValue]
- (UIBezierPath*)smoothedPathWithGranularity:(NSInteger)granularity
{
NSMutableArray *points = [(NSMutableArray*)[self pointsOrdered] mutableCopy];
if (points.count < 4) return [self bezierPath];
// Add control points to make the math make sense
[points insertObject:[points objectAtIndex:0] atIndex:0];
[points addObject:[points lastObject]];
UIBezierPath *smoothedPath = [self bezierPath];
[smoothedPath removeAllPoints];
[smoothedPath moveToPoint:POINT(0)];
for (NSUInteger index = 1; index < points.count - 2; index++)
{
CGPoint p0 = POINT(index - 1);
CGPoint p1 = POINT(index);
CGPoint p2 = POINT(index + 1);
CGPoint p3 = POINT(index + 2);
// 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];
}
// Now add p2
[smoothedPath addLineToPoint:p2];
}
// finish by adding the last point
[smoothedPath addLineToPoint:POINT(points.count - 1)];
return smoothedPath;
}
- (PVPoint *)pointAppendingCGPoint:(CGPoint)CGPoint
{
PVPoint *newPoint = [[PVPoint alloc] initInsertingIntoManagedObjectContext:[self managedObjectContext]];
[newPoint setCGPoint:CGPoint];
[newPoint setOrder:[NSNumber numberWithUnsignedInteger:[[self points] count]]];
[[self mutableSetValueForKey:@"points"] addObject:newPoint];
[(NSMutableArray *)[self pointsOrdered] addObject:newPoint];
[[self bezierPath] addLineToPoint:CGPoint];
return [newPoint autorelease];
if ([self bezierPath] && [pointsOrdered count] > 3)
{
PVPoint *control1 = [pointsOrdered objectAtIndex:[pointsOrdered count] - 2];
PVPoint *control2 = [pointsOrdered objectAtIndex:[pointsOrdered count] - 1];
[bezierPath moveToPoint:[[pointsOrdered objectAtIndex:[pointsOrdered count] - 3] CGPoint]];
[[self bezierPath] addCurveToPoint:CGPoint controlPoint1:[control1 CGPoint] controlPoint2:[control2 CGPoint]];
}
}
- (BOOL)isComplete { return [[self points] count] > 1; }
- (UIBezierPath *)bezierPath
{
if (!bezierPath)
{
bezierPath = [UIBezierPath bezierPath];
for (NSUInteger p = 0; p < [[self points] count]; p++)
{
if (!p) [bezierPath moveToPoint:[(PVPoint *)[[self pointsOrdered] objectAtIndex:p] CGPoint]];
else [bezierPath addLineToPoint:[(PVPoint *)[[self pointsOrdered] objectAtIndex:p] CGPoint]];
}
[bezierPath retain];
}
return bezierPath;
}
- (CGPathRef)CGPath
{
return [[self bezierPath] CGPath];
}
【问题讨论】：
-
你能告诉我们锯齿状的末端吗？
-
请查看上面的 UIBezierPath 代码。
-
苹果有一个名为 QuartzDemo 的示例代码
-
我知道，但是在快速绘制时它不会提供平滑曲线...我需要一个平滑公式或其他东西
-
你能发一张它现在的截图吗？
标签： iphone objective-c ios ipad bezier