That's correct, the most specialized implementation is used. In the case
where more there is no implementation more specialized than another (two
interfaces are extended to a protocol and a class implements both) then an
arbitrary implementation from the available ones will be selected.

Extending a Object and nil to a protocol is a good way to provide a default
impl.
Another option is to wrap the protocol functions in another function and
testing for satisfies?.
 Il giorno 17/nov/2014 10.49, "pmf" <phil.fr...@gmx.de> ha scritto:

> Is there a way to provide a default (fallback) implementation for a method
> defined in a defprotocol directive? I do realize that I could extend the
> protocol for type java.lang.Object, but this raises the question about how
> protocol implementations with extend are sorted, i.e. how specializations
> of types of the same hierarchy are resolved. Is the most specialized
> implementation used if I implement a protocol for a class as well as its
> parent class and if yes, is this independent of the order of definition for
> the protocol extensions?
>
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