I have the  box & lt; Specialty & gt; , and want to be able to put it on the  box < Obj & gt; . It is believed that this is to be done, but  t.downcast :: & lt; Obj & gt; Attempting to call ()  says that there is no  downcast  in the method. 
 The docs shows how to do this: If you have a reference you can just  & amp; As a specialty & amp; Anyone can .  boxedTrait as box & lt; Any & gt; It is not possible to . 
It shows what I mean.
  allows any downtake of a concrete type, so you  Required  to know the concrete type when you press the  box & lt; Any & gt;  However, you do not know the concrete type if you have only one attribute item for some other attribute - this particular property is the point of the objects then you can see the  box & lt; SomeTrait & gt;  to  box & lt; Any & gt; , it is impossible. 
 Ideally it is possible to write something like  box & lt; Show + any & gt; It will also allow the use of  any  methods with the  show  methods, it is also impossible, though: you only have additional life styles for the main feature and You can type in the underlying type, so the  box & lt; Show + Sync + 'a & gt;  is legal, but  box & lt; <+ Show any . 
 If you want to use it with  any , then one way to get it will be the property: 
  trait MyTrait: Someone Also {// ...}    However, inheritance properties do not work with objects, so you can not call   on matches  Box & lt; MyTrait & gt;   Includes reworking of any  (as can be found), but it is pretty much anything. 
Unfortunately, I'm not aware of a simple way to talk about this kind of thing possibly possible to implement with some unsafe code, but I do not know how.
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