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.
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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



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