Issue #7274 has been updated by rits (First Last).


matz (Yukihiro Matsumoto) wrote:
> @rits You haven't read my message above, have you?
> 
> >It is due to implementation limitation of CRuby.
> >
> >The structure of instances of subclass (TT_XXX) may be different from superclasses.
> >In that case, the C implemented methods would crash.
> >
> >So we prohibit them conservatively.
> 
> Matz.

Neither I nor marcandre understood what you were alluding to and asked for a clarification, but you just repeated without clarifying.

You appear to be saying, and please correct if that's wrong, that a method from a subclass can not safely be bound to an instance of a superclass.  If so, please note, that is not what's being suggested. sub_foo, from the original example, is a method that is defined in the superclass, not the subclass.

----------------------------------------
Feature #7274: UnboundMethods should be bindable  to any object that is_a?(owner of the UnboundMethod)
https://bugs.ruby-lang.org/issues/7274#change-43153

Author: rits (First Last)
Status: Rejected
Priority: Normal
Assignee: matz (Yukihiro Matsumoto)
Category: core
Target version: next minor


=begin
as a corollary, (({UnboundMethod}))s referencing the same method name on the same owner, should be equal

currently (({UnboundMethod}))s binding is determined by the class via which they were retrieved, not the owner

 class Base; def foo; end end
 class Sub < Base; end

 base_foo = Base.instance_method :foo
 sub_foo = Sub.instance_method :foo
 sub_foo.bind(Base.new).call

(({sub_foo.owner})) is (({Base})) so there does not seem to be any reason why it's not safe for it to bind to an instance of (({Base})).

and there does not seem to be any reason for (({sub_foo})) and (({base_foo})) to be unequal, they both refer to the same method, (({foo})) on (({Base})).
=end



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