From: "matz (Yukihiro Matsumoto)" <matz@...>
Date: 2013-09-18T12:22:31+09:00
Subject: [ruby-core:57253] [ruby-trunk - Feature #7274] UnboundMethods	should be bindable to any object that is_a?(owner of the	UnboundMethod)


Issue #7274 has been updated by matz (Yukihiro Matsumoto).


I am not sure the intention of the patch in [ruby-core:48864], but

* if the owner is a module
* if the binding object is an instance of the owner

we can allow bind(), but I am not sure how much useful this relaxing is, especially the latter one.

Matz.

----------------------------------------
Feature #7274: UnboundMethods should be bindable  to any object that is_a?(owner of the UnboundMethod)
https://bugs.ruby-lang.org/issues/7274#change-41872

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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