使用HSL亮度系数从SKColor返回浅色

我有一个iOS SKColor，我想转换为较浅的阴影（不透明，没有透明度/没有透明度），因此我想使用HSL亮度实现一个函数，返回一个SKColor。 即。 在函数内部，将原始SKColor转换为HSL等效值，然后将值应用于HSL亮度以获得更浅的色调，然后将此HSL颜色转换回我可以使用的SKColor。

Answer 1

Xcode 8.2.1•Swift 3.0.2

extension UIColor { convenience init(hue: CGFloat, saturation: CGFloat, lightness: CGFloat, alpha: CGFloat = 1) { let offset = saturation * (lightness < 0.5 ? lightness : 1 - lightness) let brightness = lightness + offset let saturation = lightness > 0 ? 2 * offset / brightness : 0 self.init(hue: hue, saturation: saturation, brightness: brightness, alpha: alpha) } var lighter: UIColor? { return applying(lightness: 1.25) } func applying(lightness value: CGFloat) -> UIColor? { guard let hsl = hsl else { return nil } return UIColor(hue: hsl.hue, saturation: hsl.saturation, lightness: hsl.lightness * value, alpha: hsl.alpha) } var hsl: (hue: CGFloat, saturation: CGFloat, lightness: CGFloat, alpha: CGFloat)? { var red: CGFloat = 0, green: CGFloat = 0, blue: CGFloat = 0, alpha: CGFloat = 0, hue: CGFloat = 0 guard getRed(&red, green: &green, blue: &blue, alpha: &alpha), getHue(&hue, saturation: nil, brightness: nil, alpha: nil) else { return nil } let upper = max(red, green, blue) let lower = min(red, green, blue) let range = upper - lower let lightness = (upper + lower) / 2 let saturation = range == 0 ? 0 : range / (lightness < 0.5 ? lightness * 2 : 2 - lightness * 2) return (hue, saturation, lightness, alpha) } } 

---

 let purple = UIColor(red: 160/255, green: 118/255, blue: 200/255, alpha: 1) let lighter = purple.lighter 

Answer 2

正如官方SKColor doc所说，在iOS上， SKColor只是一个UIColor 。 因此：

 UIColor *uicolor = (UIColor *)skcolor; CGFloat h, s, b, a; // lightness is called 'brightness' [uicolor getHue:&h saturation:&s brightness:&b alpha:&a]; // (play with your brightness value here) SKColor *skcolor2 = (SKColor *)[UIColor colorWithHue:h saturation:s brightness:b alpha:a];