[TIMOB-6992] iOS: UI - ActivityIndicator incorrect positioning with left and right properties

Actual Result

Setting the "left" or "right" property of a Ti.UI.ActivityIndicator yields incorrect positioning. The ActivityIndicator seems to be centering on a horizontal coordinate that is relative to the center of the view that the ActivityIndicator is added to. Examples: * ActivityIndicator with left=0 or right=0 will be centered within its view. * ActivityIndicator with left=50 will have its center positioned 50px to the right of the center of its view. * ActivityIndicator with right=-50 will have its center positioned 50px to the right of the center of its view. * ActivityIndicator with right=50 will have its center positioned 50px to the left of the center of its view.

Expected result

When the "left" or "right" property is supplied, the Ti.UI.ActivityIndicator should be positioned within the view it is added to accordingly, just like any other component would be (button, image, label, etc).

Test case

var win1 = Ti.UI.createWindow({
  backgroundColor:'#fff'
});

var label1 = Ti.UI.createLabel({
  backgroundColor:'green',
  color:'white',
  text:'Spinner should be on top of the X',
  font:{fontSize:20,fontFamily:'Helvetica Neue'},
  textAlign:'center',
  top:40,
  height:40,
  width:'auto'
});
win1.add(label1);

var view1 = Ti.UI.createView({
  backgroundColor:'yellow',
  height:30,
  top:100,
  width:100
});
win1.add(view1);

var spin1=Ti.UI.createActivityIndicator({
  //right:25,
  left:25,
  style:Ti.UI.iPhone.ActivityIndicatorStyle.DARK
});
view1.add(spin1);

var marker1=Ti.UI.createLabel({
  //right:25,
  left:25,
  height:'auto',
  width:'auto',
  text:'X'
});
view1.add(marker1);

win1.open();
spin1.show();

The code above should place the spinner right on top of the X.

Workaround

Explicitly set the activityIndicator width and height properties to 'auto'