Challenge: solve this multiple inheritance problem in your favorite language

[mw]

On Friday, 5 June 2020 at 06:40:06 UTC, Jacob Carlborg wrote:
> It's already possible to do that today:
>
> class Person : NameI, AddrI {
>   mixin NameT!Person Name;
>   mixin AddrT!Person Addr;
>
>   bool equals(Person other) {
>     return Name.equals(other) &&
>            Addr.equals(other);
>   }
> }

Thank you for letting me know.

This alleviates the name clashing problem, but didn't completely 
solve it, this renaming is still coarse-grained all-or-none, e.g:

class Visitor {
   mixin UKResident UKR;
   mixin USResident USR;
}

all the attributes in UKResident.<attr> is rename to UKR.<attr> 
and
all the attributes in USResident.<attr> is rename to USR.<attr>

While we want to achieve:

   UKR.name  === USR.name (have the same  storage)
   UKR.addr !=== USR.addr (have different storage)

i.e. fine-grained control on each mixin's attribute to be either 
joined or separated.