Calling super's super method

Question

Is is possible to do something like super.super in the overriden method? That is, to bypass the direct parent\'s super and call \"grandparent\'s\" super?

Answer 1

you could modify the method's arguments to allow some kind of optional 'pass to parent' parameter. in the super class of your child, check for this parameter and if so, call super from that method and return, otherwise allow execution to continue.

class Grandparent; def method_name(opts={}); puts "Grandparent called."; end; end
class Parent < Grandparent
def method_name(opts={})
  return super if opts[:grandparent]
  # do stuff otherwise...
  puts "Parent called."
end
end
class Child < Parent
def method_name(opts={})
  super(:grandparent=>true)
end
end

ruby-1.9.2-p0 > Child.new.method_name
Grandparent called.
  => nil   

otherwise i agree with @Femaref, just b/c something is possible doesn't mean it's a good idea. reconsider your design if you think this is necessary.

Answer 2

This is not recommended, but what you want is possible like this:

grandparent = self.class.superclass.superclass
meth = grandparent.instance_method(:the_method)
meth.bind(self).call

This works by first getting the grandparent class, then calling instance_method on it to get an UnboundMethod representing the grandparent's version of the_method. It then uses UnboundMethod#bind and Method#call to call the grandparent's method on the current object.

Answer 3

Considering this is breaking one of the principles of OOP (encapsulation), I dearly hope it isn't possible. Even the case of you trying to do this speaks of a problem with your design.