Question

The code below is where I am right now, I’ve made it so I can click a button and change from one sprite on a sprite sheet to another. The fundamentals I’m working out may be crap, I don’t really know what I’m doing! By all means, steer me straight if this is so.

What I want to know is how to loop over all the sprites in my sprite sheet. Currently I can just flip between the first and second one (only one time) via button click.

UIImageView test = new UIImageView();
test.Image = UIImage.FromFile("necromancerSpriteSheet.png");

CGImage img = test.Image.CGImage;
CALayer layer = new CALayer();

layer.Contents = img;
layer.Bounds = new RectangleF(0, 0, 128, 128);
layer.Position = new PointF(150, 250);

UIImageView theImage = new UIImageView();

layer.ContentsRect = new RectangleF(0, 0, 0.33f, 0.33f);
layer.RemoveAllAnimations();
theImage.Layer.AddSublayer(layer);

button.TouchUpInside += (s, e) => {             
    layer.ContentsRect = new RectangleF(0.33f, 0, 0.33f, 0.33f);
    layer.RemoveAllAnimations();
    theImage.Layer.AddSublayer(layer);
};

I would expect the process would be to have all the sublayers loaded/initialized then simply loop over them in some way. I haven’t had any luck figuring out how to achieve this yet though.

Here is my test spritesheet if it helps…

Answer 1

There are many ways of solving this problem, but if you want to use CoreAnimation to drive the animation on the sheet, my suggestion is that you use a keyframe animation (CAKeyFrameAnimation) to drive the animation.

To simplify things, I would merely put all of the items in a single row, and then use something like:

var anim = (CAKeyFrameAnimation) CAKeyFrameAnimation.FromKeyPath ("position.x");
anim.Values = new NSNumber [9];
var times = new NSNumber [9];
for (int i = 0; i < 9; i++){
    anim.Values [i] = NSNumber.FromFloat (i*WIDTH);
    times [i] = NSNumber.FromFloat (i/9f);
}
anim.CalculationMode = CAKeyFrameAnimation.CalculationDiscrete

You might need to tweak the animation above a little, I did the above without testing.