Create Hexagon ImageView shape in iOS

Question

I want Above Hexagon shape for my ImageView. But After implementing below code I am getting this Image

- (UIBezierPath *)roundedPolygonPathWithRect:(CGRect)square
                                    lineWidth:(CGFloat)lineWidth
                                        sides:(NSInteger)sides
                                cornerRadius:(CGFloat)cornerRadius
{
    UIBezierPath *path  = [UIBezierPath bezierPath];

    CGFloat theta       = 2.0 * M_PI / sides;                           // how much to turn at every corner
    CGFloat offset      = cornerRadius * tanf(theta / 2.0);             // offset from which to start rounding corners
    CGFloat squareWidth = MIN(square.size.width, square.size.height);   // width of the square

    // calculate the length of the sides of the polygon

    CGFloat length      = squareWidth - lineWidth;
    if (sides % 4 != 0) {                                               // if not dealing with polygon which will be square with all sides ...
        length = length * cosf(theta / 2.0) + offset/2.0;               //     ... offset it inside a circle inside the square
    }
    CGFloat sideLength = length * tanf(theta / 2.0);

    // start drawing at `point` in lower right corner

    CGFloat calc = squareWidth / 2.0 + sideLength / 2.0 - offset;

    CGPoint point = CGPointMake(calc, squareWidth - (squareWidth - length) / 2.0);
    CGFloat angle = M_PI;
    [path moveToPoint:point];

    // draw the sides and rounded corners of the polygon

    for (NSInteger side = 0; side < sides; side++)
    {
        point = CGPointMake(point.x + (sideLength - offset * 2.0) * cosf(angle), point.y + (sideLength - offset * 2.0) * sinf(angle));
        [path addLineToPoint:point];

        CGPoint center = CGPointMake(point.x + cornerRadius * cosf(angle + M_PI_2), point.y + cornerRadius * sinf(angle + M_PI_2));
        [path addArcWithCenter:center radius:cornerRadius startAngle:angle - M_PI_2 endAngle:angle + theta - M_PI_2 clockwise:YES];

        point = path.currentPoint; // we don't have to calculate where the arc ended ... UIBezierPath did that for us
        angle += theta;
    }

    [path closePath];

    return path;
}


 CGFloat lineWidth    = 5.0;
     UIBezierPath *path   = [self roundedPolygonPathWithRect:cell.eventImageView.bounds
                                              lineWidth:lineWidth
                                                  sides:6
                                           cornerRadius:10];

     CAShapeLayer *mask   = [CAShapeLayer layer];
     mask.path            = path.CGPath;
     mask.lineWidth       = lineWidth;
     mask.strokeColor     = [UIColor clearColor].CGColor;
     mask.fillColor       = [UIColor whiteColor].CGColor;
     cell.eventImageView.layer.mask = mask;

     CAShapeLayer *border = [CAShapeLayer layer];
     border.path          = path.CGPath;
     border.lineWidth     = lineWidth;
     border.strokeColor   = [UIColor blackColor].CGColor;
     border.fillColor     = [UIColor clearColor].CGColor;
     [cell.eventImageView.layer addSublayer:border];

Please help me how can i implement this and I have never used Bezier Paths before.

Thanks in Advance !

Answer 1

I refactored the path creation function in Swift to also take a rotationOffset argument, allowing to arbitrarily rotate the regular polygon.

I am not completely sure my function is equivalent to the one you have (as I use polar coordinates to draw the polygon), but the produced result looks similar to what you want.

public func roundedPolygonPath(rect: CGRect, lineWidth: CGFloat, sides: NSInteger, cornerRadius: CGFloat, rotationOffset: CGFloat = 0) -> UIBezierPath {
    let path = UIBezierPath()
    let theta: CGFloat = CGFloat(2.0 * M_PI) / CGFloat(sides) // How much to turn at every corner
    let offset: CGFloat = cornerRadius * tan(theta / 2.0)     // Offset from which to start rounding corners
    let width = min(rect.size.width, rect.size.height)        // Width of the square

    let center = CGPoint(x: rect.origin.x + width / 2.0, y: rect.origin.y + width / 2.0)

    // Radius of the circle that encircles the polygon
    // Notice that the radius is adjusted for the corners, that way the largest outer
    // dimension of the resulting shape is always exactly the width - linewidth
    let radius = (width - lineWidth + cornerRadius - (cos(theta) * cornerRadius)) / 2.0

    // Start drawing at a point, which by default is at the right hand edge
    // but can be offset
    var angle = CGFloat(rotationOffset)

    let corner = CGPointMake(center.x + (radius - cornerRadius) * cos(angle), center.y + (radius - cornerRadius) * sin(angle))
    path.moveToPoint(CGPointMake(corner.x + cornerRadius * cos(angle + theta), corner.y + cornerRadius * sin(angle + theta)))

    for _ in 0..<sides {
        angle += theta

        let corner = CGPointMake(center.x + (radius - cornerRadius) * cos(angle), center.y + (radius - cornerRadius) * sin(angle))
        let tip = CGPointMake(center.x + radius * cos(angle), center.y + radius * sin(angle))
        let start = CGPointMake(corner.x + cornerRadius * cos(angle - theta), corner.y + cornerRadius * sin(angle - theta))
        let end = CGPointMake(corner.x + cornerRadius * cos(angle + theta), corner.y + cornerRadius * sin(angle + theta))

        path.addLineToPoint(start)
        path.addQuadCurveToPoint(end, controlPoint: tip)
    }

    path.closePath()

    // Move the path to the correct origins
    let bounds = path.bounds
    let transform = CGAffineTransformMakeTranslation(-bounds.origin.x + rect.origin.x + lineWidth / 2.0, -bounds.origin.y + rect.origin.y + lineWidth / 2.0)
    path.applyTransform(transform)

    return path
}

For example, with rotationOffset set to M_PI / 6.0, the produced shape will look something like this

Just in case, you can see the full playground I used here

UPDATE (March 14 2018): Updated the gist for Swift 4 syntax, can be seen here.