ruby variable as same object (pointers?)

Question

>> a = 5
=> 5
>> b = a
=> 5
>> b = 4
=> 4
>> a
=> 5

how can I set \'b\' to actually be \'a\' so that in the exa

Answer 1

As has been noted the syntax you are using can not be done. Just throwing this out there though you could make a wrapper class it depends what you actually want to do

ruby-1.8.7-p334 :007 > class Wrapper
ruby-1.8.7-p334 :008?>   attr_accessor :number
ruby-1.8.7-p334 :009?>   def initialize(number)
ruby-1.8.7-p334 :010?>     @number = number
ruby-1.8.7-p334 :011?>   end
ruby-1.8.7-p334 :012?> end
 => nil 
ruby-1.8.7-p334 :013 > a = Wrapper.new(4)
 => # 
ruby-1.8.7-p334 :014 > b = a
 => # 
ruby-1.8.7-p334 :015 > a.number = 6
 => 6 
ruby-1.8.7-p334 :016 > a
 => # 
ruby-1.8.7-p334 :017 > b
 => #