38

我需要在我的 Draw 应用程序中用箭头画线。我三角学不好,所以解决不了这个问题。

用户将手指放在屏幕上并在任何方向上画线。所以,箭头应该出现在行尾。

4

4 回答 4

198

更新

我已经单独发布了这个答案的 Swift 版本。

原来的

这是一个有趣的小问题。首先,有很多方法来绘制箭头,有弯曲的或直的。让我们选择一种非常简单的方法并标记我们需要的测量值:

箭头零件

我们想写一个函数,它接受起点、终点、尾宽、头宽和头长,并返回一个勾勒出箭头形状的路径。让我们创建一个名为的类别以dqd_arrowhead将此方法添加到UIBezierPath

// UIBezierPath+dqd_arrowhead.h

@interface UIBezierPath (dqd_arrowhead)

+ (UIBezierPath *)dqd_bezierPathWithArrowFromPoint:(CGPoint)startPoint
                                           toPoint:(CGPoint)endPoint
                                         tailWidth:(CGFloat)tailWidth
                                         headWidth:(CGFloat)headWidth
                                        headLength:(CGFloat)headLength;

@end

由于箭头的路径上有七个角,让我们通过命名该常量来开始我们的实现:

// UIBezierPath+dqd_arrowhead.m

#import "UIBezierPath+dqd_arrowhead.h"

#define kArrowPointCount 7

@implementation UIBezierPath (dqd_arrowhead)

+ (UIBezierPath *)dqd_bezierPathWithArrowFromPoint:(CGPoint)startPoint
                                           toPoint:(CGPoint)endPoint
                                         tailWidth:(CGFloat)tailWidth
                                         headWidth:(CGFloat)headWidth
                                        headLength:(CGFloat)headLength {

OK,简单的部分就完成了。现在,我们如何找到路径上这七个点的坐标?如果箭头沿 X 轴对齐,则更容易找到点:

轴对齐的箭头点

计算轴对齐箭头上的点坐标非常容易,但我们需要箭头的总长度来计算。我们将使用hypotf标准库中的函数:

    CGFloat length = hypotf(endPoint.x - startPoint.x, endPoint.y - startPoint.y);

我们将调用一个辅助方法来实际计算这七个点:

    CGPoint points[kArrowPointCount];
    [self dqd_getAxisAlignedArrowPoints:points
                              forLength:length
                              tailWidth:tailWidth
                              headWidth:headWidth
                             headLength:headLength];

但是我们需要转换这些点,因为通常我们不会尝试创建轴对齐的箭头。幸运的是,Core Graphics 支持一种称为仿射变换的变换,它可以让我们旋转和平移(滑动)点。我们将调用另一个辅助方法来创建将轴对齐的箭头转换为我们被要求的箭头的转换:

    CGAffineTransform transform = [self dqd_transformForStartPoint:startPoint
                                                          endPoint:endPoint
                                                            length:length];

现在我们可以使用轴对齐箭头的点和将其转换为我们想要的箭头的变换来创建 Core Graphics 路径:

    CGMutablePathRef cgPath = CGPathCreateMutable();
    CGPathAddLines(cgPath, &transform, points, sizeof points / sizeof *points);
    CGPathCloseSubpath(cgPath);

最后,我们可以将 a 包裹UIBezierPath起来CGPath并返回它:

    UIBezierPath *uiPath = [UIBezierPath bezierPathWithCGPath:cgPath];
    CGPathRelease(cgPath);
    return uiPath;
}

这是计算点坐标的辅助方法。这很简单。如果需要,请参阅轴对齐箭头的图表。

+ (void)dqd_getAxisAlignedArrowPoints:(CGPoint[kArrowPointCount])points
                            forLength:(CGFloat)length
                            tailWidth:(CGFloat)tailWidth
                            headWidth:(CGFloat)headWidth
                           headLength:(CGFloat)headLength {
    CGFloat tailLength = length - headLength;
    points[0] = CGPointMake(0, tailWidth / 2);
    points[1] = CGPointMake(tailLength, tailWidth / 2);
    points[2] = CGPointMake(tailLength, headWidth / 2);
    points[3] = CGPointMake(length, 0);
    points[4] = CGPointMake(tailLength, -headWidth / 2);
    points[5] = CGPointMake(tailLength, -tailWidth / 2);
    points[6] = CGPointMake(0, -tailWidth / 2);
}

计算仿射变换更复杂。这就是三角函数的用武之地。您可以使用atan2andCGAffineTransformRotateCGAffineTransformTranslate函数来创建它,但是如果您记住了足够的三角函数,则可以直接创建它。请参阅Quartz 2D 编程指南中的“矩阵背后的数学”,以获取有关我在这里所做工作的更多信息:

+ (CGAffineTransform)dqd_transformForStartPoint:(CGPoint)startPoint
                                       endPoint:(CGPoint)endPoint
                                         length:(CGFloat)length {
    CGFloat cosine = (endPoint.x - startPoint.x) / length;
    CGFloat sine = (endPoint.y - startPoint.y) / length;
    return (CGAffineTransform){ cosine, sine, -sine, cosine, startPoint.x, startPoint.y };
}

@end

我已将所有代码放在一个要点中,以便轻松复制'n'paste

使用此类别,您可以轻松绘制箭头:

示例箭头 1 示例箭头 2

由于您只是在生成路径,因此您可以选择不填充它,或者不描边它,如下例所示:

未描边的箭头样本

不过,你必须小心。如果您使头部宽度小于尾部宽度,或者如果您使头部长度大于总箭头长度,则此代码不会阻止您获得时髦的结果:

窄头样品 头部过长样本

于 2012-11-26T06:01:51.490 回答
18

这是我的旧 Objective-C 代码的 Swift 版本。它应该在 Swift 3.2 和更高版本中工作。

extension UIBezierPath {

    static func arrow(from start: CGPoint, to end: CGPoint, tailWidth: CGFloat, headWidth: CGFloat, headLength: CGFloat) -> UIBezierPath {
        let length = hypot(end.x - start.x, end.y - start.y)
        let tailLength = length - headLength

        func p(_ x: CGFloat, _ y: CGFloat) -> CGPoint { return CGPoint(x: x, y: y) }
        let points: [CGPoint] = [
            p(0, tailWidth / 2),
            p(tailLength, tailWidth / 2),
            p(tailLength, headWidth / 2),
            p(length, 0),
            p(tailLength, -headWidth / 2),
            p(tailLength, -tailWidth / 2),
            p(0, -tailWidth / 2)
        ]

        let cosine = (end.x - start.x) / length
        let sine = (end.y - start.y) / length
        let transform = CGAffineTransform(a: cosine, b: sine, c: -sine, d: cosine, tx: start.x, ty: start.y)

        let path = CGMutablePath()
        path.addLines(between: points, transform: transform)
        path.closeSubpath()

        return self.init(cgPath: path)
    }

}

这是您如何称呼它的示例:

let arrow = UIBezierPath.arrow(from: CGPoint(x: 50, y: 100), to: CGPoint(x: 200, y: 50),
        tailWidth: 10, headWidth: 25, headLength: 40)
于 2016-06-23T08:06:58.777 回答
8
//This is the integration into the view of the previous exemple
//Attach the following class to your view in the xib file

#import <UIKit/UIKit.h>

@interface Arrow : UIView

@end

#import "Arrow.h"
#import "UIBezierPath+dqd_arrowhead.h"

@implementation Arrow
{
    CGPoint startPoint;
    CGPoint endPoint;
    CGFloat tailWidth;
    CGFloat headWidth;
    CGFloat headLength;
    UIBezierPath *path;

}


- (id)initWithCoder:(NSCoder *)aDecoder
{
    if (self = [super initWithCoder:aDecoder])
    {
        [self setMultipleTouchEnabled:NO];
        [self setBackgroundColor:[UIColor whiteColor]];

    }
    return self;
}

- (void)drawRect:(CGRect)rect {

    [[UIColor redColor] setStroke];
    tailWidth = 4;
    headWidth = 8;
    headLength = 8;
    path = [UIBezierPath dqd_bezierPathWithArrowFromPoint:(CGPoint)startPoint
                                                  toPoint:(CGPoint)endPoint
                                                tailWidth:(CGFloat)tailWidth
                                                headWidth:(CGFloat)headWidth
                                               headLength:(CGFloat)headLength];
    [path setLineWidth:2.0];

    [path stroke];

}
- (void) touchesBegan:(NSSet*)touches withEvent:(UIEvent*)event
{
    UITouch* touchPoint = [touches anyObject];
    startPoint = [touchPoint locationInView:self];
    endPoint = [touchPoint locationInView:self];


    [self setNeedsDisplay];
}

-(void)touchesEnded:(NSSet *)touches withEvent:(UIEvent *)event
{
    UITouch* touch = [touches anyObject];
    endPoint=[touch locationInView:self];
    [self setNeedsDisplay];
}

-(void)touchesMoved:(NSSet *)touches withEvent:(UIEvent *)event
{
    UITouch* touch = [touches anyObject];
    endPoint = [touch locationInView:self];
    [self setNeedsDisplay];
}


@end
于 2013-06-18T12:21:23.860 回答
2

在 Swift 3.0 中,您可以使用

extension UIBezierPath {

class func arrow(from start: CGPoint, to end: CGPoint, tailWidth: CGFloat, headWidth: CGFloat, headLength: CGFloat) -> Self {
    let length = hypot(end.x - start.x, end.y - start.y)
    let tailLength = length - headLength

    func p(_ x: CGFloat, _ y: CGFloat) -> CGPoint { return CGPoint(x: x, y: y) }
    var points: [CGPoint] = [
        p(0, tailWidth / 2),
        p(tailLength, tailWidth / 2),
        p(tailLength, headWidth / 2),
        p(length, 0),
        p(tailLength, -headWidth / 2),
        p(tailLength, -tailWidth / 2),
        p(0, -tailWidth / 2)
    ]

    let cosine = (end.x - start.x) / length
    let sine = (end.y - start.y) / length
    var transform = CGAffineTransform(a: cosine, b: sine, c: -sine, d: cosine, tx: start.x, ty: start.y)        
    let path = CGMutablePath()
    path.addLines(between: points, transform: transform)
    path.closeSubpath()
    return self.init(cgPath: path)
}

}
于 2017-01-06T13:35:54.660 回答