找点接近，该线的端点之间(Finding point is close to line and be

找到，如果该点位于包含两点我做以下检查指定行：

-(Boolean)isOnLine:(Line*) line point:(CGPoint) point{
    //If between two dots:
    if (((line.first.x <= point.x && point.x <= line.last.x)||(line.first.x >= point.x && point.x >= line.last.x))&&((line.first.y<=point.y && point.y<= line.last.y)||(line.first.y>=point.y && point.y>=line.last.y)) ) {    
        //Calculate distance:
        double dist = (((double)point.y - line.first.y)) / (0.00001+((double)(point.x - line.first.x)))- ((double)(line.last.y - line.first.y)) / (0.00001+((double)(line.last.x - line.first.x)));
        NSLog(@"Dist to line: %f", fabs(dist));
        return fabs(dist) <0.5;
    }else
        return NO;  
    }
}

不知何故，但是，功能不与垂直线工作。我的猜测是if从句在某种意义上是无效的。

Answer 1:

我没有看到你的代码仔细，所以我不完全知道你在做什么，但FYI做此操作的最简单方法是找到行的点的一端的距离，发现对方的距离行至点的结束，然后添加这些距离，并比较该线的长度。

就像是：

Boolean isOnLine(line, point) {
  var dist1 = dist(line.first, point)
  var dist2 = dist(line.last, point)
  return abs(line.length - (dist1 + dist2)) < .5
}

对于DIST（）函数，我猜CoreGraphics中规定，但如果没有它只是基本的三角。

Answer 2:

下面是我实现jhockings的解决方案

    return abs([line length] -
    (sqrt((line.first.x - point.x)*(line.first.x - point.x)
    + (line.first.y - point.y)*(line.first.y - point.y))
    + sqrt((line.last.x - point.x)*(line.last.x - point.x)
    + (line.last.y - point.y)*(line.last.y - point.y)))) < .5;

Answer 3:

另一个（我）实现@jhocking解决方案：

- (BOOL)isPoint:(CGPoint)origin nearToLineSegmentPointA:(CGPoint)pointA pointB:(CGPoint)pointB withMarginOfError:(CGFloat)marginOfError {
    CGFloat distanceAP = [self distanceBetweenPointA:origin pointB:pointA];
    CGFloat distanceBP = [self distanceBetweenPointA:origin pointB:pointB];
    CGFloat distanceAB = [self distanceBetweenPointA:pointA pointB:pointB];

    if (fabsf(distanceAB - distanceAP - distanceBP) < marginOfError) {
        return YES;
    } else {
        return NO;
    }
}

- (CGFloat)distanceBetweenPointA:(CGPoint)pointA pointB:(CGPoint)pointB {
    return sqrtf(powf((pointA.x - pointB.x), 2.f) + powf((pointA.y - pointB.y), 2.f));
}