calling another method in super class in ruby

Question

class B < A

  alias :super_a :a

  def a
    b()
  end
  def b
    super_a()
  end
end  

There's no nice way to do it, but you can do A.instance_method(:a).bind(self).call, which will work, but is ugly.

You could even define your own method in Object to act like super in java:

class SuperProxy
  def initialize(obj)
    @obj = obj
  end

  def method_missing(meth, *args, &blk)
    @obj.class.superclass.instance_method(meth).bind(@obj).call(*args, &blk)
  end
end

class Object
  private
  def sup
    SuperProxy.new(self)
  end
end

class A
  def a
    puts "In A#a"
  end
end

class B<A
  def a
  end

  def b
    sup.a
  end
end
B.new.b # Prints in A#a

If you don't explicitly need to call A#a from B#b, but rather need to call A#a from B#a, which is effectively what you're doing by way of B#b (unless you're example isn't complete enough to demonstrate why you're calling from B#b, you can just call super from within B#a, just like is sometimes done in initialize methods. I know this is kind of obvious, I just wanted to clarify for any Ruby new-comers that you don't have to alias (specifically this is sometimes called an "around alias") in every case.

class A
  def a
    # do stuff for A
  end
end

class B < A
  def a
    # do some stuff specific to B
    super
    # or use super() if you don't want super to pass on any args that method a might have had
    # super/super() can also be called first
    # it should be noted that some design patterns call for avoiding this construct
    # as it creates a tight coupling between the classes.  If you control both
    # classes, it's not as big a deal, but if the superclass is outside your control
    # it could change, w/o you knowing.  This is pretty much composition vs inheritance
  end
end