问题描述
我想使用我从 MyView hex = new MyView(this,yellow,10,90);
doodleView.addView(hex);
Path hexPath = new Path();
hexPath.moveto(10,10);
hexPath.lineto(10,PHONE_Dims.y *4/5);
// you've already moved to this position with the above lineto() call
// hexPath.moveto(10,PHONE_Dims.y *4/5);
// Animate the x and y properties,not startX and startY
ObjectAnimator hexAn = ObjectAnimator.ofFloat(hex,"x","y",hexPath);
OvershootInterpolator interpolator = new OvershootInterpolator();
hexAn.setInterpolator(interpolator);
hexAn.setDuration(5000);
hexAn.start();
得到的 Class<T> 并用它在类中构建一个类型为 T 的变量。我不太确定如何描述它，所以这里有一些代码，希望能够很好地描述它。
Type.resolveClass()所以我想知道如何从 public static function build(name:String) {
var fields = Context.getBuildFields();
// Get Class<T>
var resolved = Type.resolveClass(name);
// Do some magic to get a ComplexType that represents the T in Class<T>
var type = magic(resolved); // I want to find out how to do this
// Create a field of type T
fields.push({
name: "ofTypeName",pos: Context.currentPos(),access: [AStatic,APublic],kind: FVar(type,macro null)
});
return fields;
}
中得到一个 ComplexType 代表 Class<T>
解决方法
var type = Context.getType(name);
var ctype = TypeTools.toComplexType(type);