From: "rits (First Last)" Date: 2013-11-26T14:04:08+09:00 Subject: [ruby-core:58583] [ruby-trunk - Feature #7274] UnboundMethods should be bindable to any object that is_a?(owner of the UnboundMethod) Issue #7274 has been updated by rits (First Last). to continue, the above: we are binding a method from Base to an instance of Base, and it's failing. Why? How can that possibly be unsafe? What difference does it make that the method was requested from a subclass of Base? It certainly does not change the fact that the method is from Base. ---------------------------------------- Feature #7274: UnboundMethods should be bindable to any object that is_a?(owner of the UnboundMethod) https://bugs.ruby-lang.org/issues/7274#change-43154 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 -- http://bugs.ruby-lang.org/