如何从UIBezierPath获取点列表？

Question

我有一个UIBezierPath,我需要从中获取一个点列表.

在Qt有一个函数被称为pointAtPercent适合我的需要,但我找不到相应的东西Objective-C.

有谁知道如何做到这一点？

Answer 1

你可以试试这个:

UIBezierPath *yourPath; // Assume this has some points in it
CGPath yourCGPath = yourPath.CGPath;
NSMutableArray *bezierPoints = [NSMutableArray array];
CGPathApply(yourCGPath, bezierPoints, MyCGPathApplierFunc);

路径应用程序功能将依次传递每个路径的路径元素.
查看CGPathApplierFunction和CGPathApply.

路径应用程序功能可能看起来像这样

void MyCGPathApplierFunc (void *info, const CGPathElement *element) {
    NSMutableArray *bezierPoints = (NSMutableArray *)info;

    CGPoint *points = element->points;
    CGPathElementType type = element->type;

    switch(type) {
        case kCGPathElementMoveToPoint: // contains 1 point
            [bezierPoints addObject:[NSValue valueWithCGPoint:points[0]]];
            break;

        case kCGPathElementAddLineToPoint: // contains 1 point
            [bezierPoints addObject:[NSValue valueWithCGPoint:points[0]]];            
            break;

        case kCGPathElementAddQuadCurveToPoint: // contains 2 points
            [bezierPoints addObject:[NSValue valueWithCGPoint:points[0]]];
            [bezierPoints addObject:[NSValue valueWithCGPoint:points[1]]];            
            break;

        case kCGPathElementAddCurveToPoint: // contains 3 points
            [bezierPoints addObject:[NSValue valueWithCGPoint:points[0]]];
            [bezierPoints addObject:[NSValue valueWithCGPoint:points[1]]];
            [bezierPoints addObject:[NSValue valueWithCGPoint:points[2]]];
            break;

        case kCGPathElementCloseSubpath: // contains no point
            break;
    }
}

Answer 2

@Moritz很棒的回答!为了找到类似Swift的解决方案,我在这篇文章中遇到了一种方法并实现了这样的CGPath扩展来从路径中获取点:

extension CGPath {
    func points() -> [CGPoint]
    {
        var bezierPoints = [CGPoint]()
        self.forEach({ (element: CGPathElement) in
            let numberOfPoints: Int = {
                switch element.type {
                case .MoveToPoint, .AddLineToPoint: // contains 1 point
                    return 1
                case .AddQuadCurveToPoint: // contains 2 points
                    return 2
                case .AddCurveToPoint: // contains 3 points
                    return 3
                case .CloseSubpath:
                    return 0
                }
            }()
            for index in 0..<numberOfPoints {
                let point = element.points[index]
                bezierPoints.append(point)
            }
        })
        return bezierPoints
    }

    func forEach(@noescape body: @convention(block) (CGPathElement) -> Void) {
        typealias Body = @convention(block) (CGPathElement) -> Void
        func callback(info: UnsafeMutablePointer<Void>, element: UnsafePointer<CGPathElement>) {
            let body = unsafeBitCast(info, Body.self)
            body(element.memory)
        }
        let unsafeBody = unsafeBitCast(body, UnsafeMutablePointer<Void>.self)
        CGPathApply(self, unsafeBody, callback)
    }
}

Answer 3

我想你试图做这样的事情:

https://math.stackexchange.com/questions/26846/is-there-an-explicit-form-for-cubic-bézier-curves

y=u0(1?x^3)+3u1(1?x^2)x+3u2(1?x)x^2+u3x^3

这是我打印该函数的所有值的方法.

- (void)logXY {

    float u0 = 0;
    float u1 = 0.05;
    float u2 = 0.25;
    float u3 = 1;

    for (float x = 0; x <= 10.0; x = x + 0.1) {
        float y = u0 * (1 - x * x * x) + 3 * u1 * (1 - x * x) * x + 3 * u2 * (1 - x) * x * x + u3 * x * x * x;

        NSLog(@"x: %f\ty: %f", x, y);
    }
}

输出是:

x: 0.000000 y: 0.000000
x: 0.100000 y: 0.022600
x: 0.200000 y: 0.060800
x: 0.300000 y: 0.115200
x: 0.400000 y: 0.186400
x: 0.500000 y: 0.275000
x: 0.600000 y: 0.381600
x: 0.700000 y: 0.506800
x: 0.800000 y: 0.651200
x: 0.900000 y: 0.815400
x: 1.000000 y: 1.000000
x: 1.100000 y: 1.205600
x: 1.200000 y: 1.432800
x: 1.300000 y: 1.682200
x: 1.400000 y: 1.954401
x: 1.500000 y: 2.250001
x: 1.600000 y: 2.569601
x: 1.700000 y: 2.913801
x: 1.800000 y: 3.283201
x: 1.900000 y: 3.678401
x: 2.000000 y: 4.100001
x: 2.100000 y: 4.548601
x: 2.200000 y: 5.024800
x: 2.300000 y: 5.529200
x: 2.400000 y: 6.062399
x: 2.500000 y: 6.625000
x: 2.600000 y: 7.217597
x: 2.700000 y: 7.840797
x: 2.799999 y: 8.495197
x: 2.899999 y: 9.181394
x: 2.999999 y: 9.899996