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    Home»iOS Development»Rounded corner of a cut-out arc using UIBezierPath
    iOS Development

    Rounded corner of a cut-out arc using UIBezierPath

    AdminBy AdminDecember 12, 2025No Comments2 Mins Read2 Views
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    Rounded corner of a cut-out arc using UIBezierPath
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    I’m drawing a cut-out hole in CAShapeLayer using UIBezierPath, this is the code I have so far:

    let view = UIView(
      frame: CGRect(origin: .zero, size: CGSize(width: 500.0, height: 500.0))
    )
    view.backgroundColor = .clear
    
    let shapeLayer = CAShapeLayer()
    shapeLayer.frame = view.bounds
    shapeLayer.fillColor = UIColor.systemMint.cgColor
    
    let mainPath = UIBezierPath(
      roundedRect: CGRect(
        origin: CGPoint(x: 50.0, y: 50.0), size: CGSize(width: 400.0, height: 400.0)
      ),
      cornerRadius: 10.0
    )
    
    let cutOutArc = UIBezierPath()
    cutOutArc.addArc(
      withCenter: CGPoint(x: 50.0, y: 250.0),
      radius: 50.0,
      startAngle: .pi / 2.0,
      endAngle: .pi + .pi / 2.0,
      clockwise: false
    )
    cutOutArc.addLine(to: CGPoint(x: 50.0, y: 200.0))
    cutOutArc.close()
    
    mainPath.append(cutOutArc)
    
    shapeLayer.fillRule = .evenOdd
    shapeLayer.path = mainPath.cgPath
    
    view.layer.insertSublayer(shapeLayer, at: 0)
    

    (The code is simplified and adapted to use in Playground.)

    The code results in this shape (looks exactly as I need):

    Rounded corner of a cut-out arc using UIBezierPath

    What I can’t figure out how to do is adding a corner radius to the cut-out arc (where the red arrows in the image below point), say, of 4.0 points. I’ve played with pretty random options for a couple days now with no meaningful result. I tried to ask AI, and it couldn’t figure out it either (it just kept on producing broken code really).

    A square shape with a half-arc cut-out hole in the centre of the left edge, and red arrows pointing to the place, where I need to add corner radius

    Does anyone know how to do it? Any guidance is much appreciated!



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