Swift Double is Not Convertible to CGFloat
Convert the values that need to be CGFloat to a CGFloat.
path.addArcWithCenter(center, radius: CGFloat(radius), startAngle: CGFloat(0.0), endAngle: CGFloat(M_PI) * 2.0, clockwise: true)startAngle probably shouldn't need to be converted though if you're just passing a literal. Also note that this isn't a C style cast, but actually converting between different Swift Types.
Edit: Looking at your whole function, this works.
func drawCircle() { // Drawing code var bounds:CGRect = self.view.bounds var center = CGPoint() center.x = bounds.origin.x + bounds.size.width / 2.0 center.y = bounds.origin.y + bounds.size.height / 2.0 var radius = (min(bounds.size.width, bounds.size.height) / 2.0) var path:UIBezierPath = UIBezierPath() path.addArcWithCenter(center, radius: CGFloat(radius), startAngle: CGFloat(0.0), endAngle: CGFloat(Float(M_PI) * 2.0), clockwise: true) path.stroke() }
You must type cast it via CGFloat(0.0). CGFloat has been adjusted to evaluate differently throughout the beta version of Xcode 6 due to the fact that in Obj-C, CGFloat casts to either a float or a double depending on the target (64 bit versus 32 bit). You must type cast a number to CGFloat in Swift to use a CGFloat as you're never guaranteed to have a float or a double (because this is dependent on the environment). This way, Swift won't throw a fit and will still be 'type' safe.