Apply Gradient to dashed UIBezierPath - ios

Question

This gradient is what I'd like to achieve on dashed UIBezierPath.
Tried this code with no success.
let gradient = CAGradientLayer()
gradient.frame = outerPath.bounds
gradient.colors = [UIColor.red.cgColor, UIColor.yellow.cgColor]
// mask
let shapeMask = CAShapeLayer()
shapeMask.path = outerPath.cgPath
self.insertSublayer(gradient, at: 0)

Answer 1

Let the CAShapeLayer make a dashed line mask.
class
TheView: UIView {
required init?( coder aDecoder: NSCoder ) {
super.init( coder: aDecoder )
let outerPath = UIBezierPath( ovalIn: bounds.insetBy( dx: 20, dy: 20 ) )
let gradient = CAGradientLayer()
gradient.frame = bounds
gradient.colors = [ UIColor.red.cgColor, UIColor.yellow.cgColor ]
let shapeMask = CAShapeLayer()
shapeMask.path = outerPath.cgPath
shapeMask.lineWidth = 8
shapeMask.lineDashPhase = 0
shapeMask.lineDashPattern = [ 6, 2 ]
shapeMask.lineCap = .butt
shapeMask.lineJoin = .bevel
shapeMask.strokeColor = UIColor.black.cgColor // Any color
shapeMask.fillColor = nil
gradient.mask = shapeMask
layer.addSublayer( gradient )
}
}

Answer 2

I think the best way is filling every square's color by yourself.In this case, the start color is green, end color is red, so how to calculate the internal color?
You can see this demo:
typedef struct{
float red;
float green;
float blue;
} TFColor;
- (void)viewDidLoad {
[super viewDidLoad];
TFColor start = {0,1,0}; //green
TFColor end = {1,0,0}; //red
for (int i = 0; i<20; i++) {
UIView *square = [[UIView alloc] initWithFrame:CGRectMake(150, 20*i, 75, 20)];
//the lerp rate changes from 1 to 0.
TFColor lerpColor = [self lerp:(19-i)/19.0f first:start second:end];
square.backgroundColor = [UIColor colorWithRed:lerpColor.red green:lerpColor.green blue:lerpColor.blue alpha:1];
[self.view addSubview:square];
}
}
#define lerpNum(a,b,r) (a*r+(1.0f-r)*b)
-(TFColor)lerp:(CGFloat)rate first:(TFColor)firstColor second:(TFColor)secondColor{
TFColor result;
result.red = lerpNum(firstColor.red, secondColor.red, rate);
result.green = lerpNum(firstColor.green, secondColor.green, rate);
result.blue = lerpNum(firstColor.blue, secondColor.blue, rate);
return result;
}
You can use the angle of every square to get smooth lerp rate.