修剪CGRect中的直线（2 CGPoint）

Question

给定CGRect并用2创建的线CGPoint是否有找到直线与直线相交的坐标的方法？

从前面的图像：我想要做到的，是替代点的矩形之外与相交的矩形边框的红点。

简而言之，我正在寻找一种方法来修剪矩形内的一条线。

这是一个数学问题，但我想知道如何使用基础解决这个问题，如果可能的话。

以下最新评论：似乎Core Graphics在这个过程中不能真正有用。我可以在Swift中转换的任何其他提示或公式？

Answer 1

像这样的东西（随便测试），基于How do you detect where two line segments intersect?

import CoreGraphics 

let rect = CGRect(x: 10, y: 10, width: 100, height: 100) 

let point1 = CGPoint(x: 200, y: 200) 
let point2 = CGPoint(x: 20, y: 20) 

struct LineSegment { 
    var point1: CGPoint 
    var point2: CGPoint 

    func intersection(with line: LineSegment) -> CGPoint? { 
     // We'll use Gavin's interpretation of LeMothe: 
     // https://stackoverflow.com/a/1968345/97337 

     let p0_x = self.point1.x 
     let p0_y = self.point1.y 
     let p1_x = self.point2.x 
     let p1_y = self.point2.y 

     let p2_x = line.point1.x 
     let p2_y = line.point1.y 
     let p3_x = line.point2.x 
     let p3_y = line.point2.y 

     let s1_x = p1_x - p0_x 
     let s1_y = p1_y - p0_y 
     let s2_x = p3_x - p2_x 
     let s2_y = p3_y - p2_y 

     let denom = (-s2_x * s1_y + s1_x * s2_y) 

     // Make sure the lines aren't parallel 
     guard denom != 0 else { return nil } // parallel 

     let s = (-s1_y * (p0_x - p2_x) + s1_x * (p0_y - p2_y))/denom 
     let t = (s2_x * (p0_y - p2_y) - s2_y * (p0_x - p2_x))/denom 

     // We've parameterized these lines as "origin + scale*vector" 
     // (s is the "scale" along one line, t is the "scale" along the other. 
     // At scale=0, we're at the origin at scale=1, we're at the terminus. 
     // Make sure we crossed between those. For more on what I mean by 
     // "parameterized" and why we go from 0 to 1, look up Bezier curves. 
     // We're just making a 1-dimentional Bezier here. 
     guard (0...1).contains(s) && (0...1).contains(t) else { return nil } 

     // Collision detected 
     return CGPoint(x: p0_x + (t * s1_x), y: p0_y + (t * s1_y)) 
    } 
} 

extension CGRect {  
    var edges: [LineSegment] { 
     return [ 
      LineSegment(point1: CGPoint(x: minX, y: minY), point2: CGPoint(x: minX, y: maxY)), 
      LineSegment(point1: CGPoint(x: minX, y: minY), point2: CGPoint(x: maxX, y: minY)), 
      LineSegment(point1: CGPoint(x: minX, y: maxY), point2: CGPoint(x: maxX, y: maxY)), 
      LineSegment(point1: CGPoint(x: maxX, y: minY), point2: CGPoint(x: maxX, y: maxY)), 
     ] 
    } 

    func intersection(with line: LineSegment) -> CGPoint? { 

    // Let's be super-simple here and require that one point be in the box and one point be outside, 
    // then we can ignore lots of corner cases 
     guard contains(line.point1) && !contains(line.point2) || 
      contains(line.point2) && !contains(line.point1) else { return nil } 

     // There are four edges. We might intersect with any of them (we know 
     // we intersect with exactly one, based on the previous guard. 
     // We could do a little math and figure out which one it has to be, 
     // but the `if` would be really tedious, so let's just check them all. 
     for edge in edges { 
      if let p = edge.intersection(with: line) { 
       return p 
      } 
     } 

     return nil 
    } 
} 

rect.intersection(with: LineSegment(point1: point1, point2: point2))