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Improve my bland List

Name: Anonymous 2012-02-04 2:25

class List
        attr_reader :car, :cdr, :size
        def initialize(*args)
                if args.size.zero? then
                        @car = nil; @cdr = nil; @size = 0
                elsif args.size == 1
                        @car = args[0]; @cdr = List.new; @size = 1
                else
                        @car = args[0]
                        @cdr = List.new *args[1 .. args.size - 1]
                        @size = args.size
                end
        end
        def [](n)
                raise ArgumentError unless n.class == Fixnum
                unless n > @size - 1 then
                        if n.zero? then @car
                        else @cdr[n - 1] end
                end
        end
        def []=(n,a)
                raise ArgumentError unless n.class == Fixnum
                unless n > @size - 1 then
                        if n.zero? then @car = a
                        else @cdr[n-1] = a end
                else raise ArgumentError
                end
        end
        def length
                @size
        end
        def null?
                @cdr.nil?
        end
        def member?(a)
                if a.nil? then true end
                unless self.null? then
                        if @car == a then true
                        else @cdr.member? a end
                else false
                end
        end
        def include?(a)
                self.member? a
        end
end

Name: Anonymous 2012-02-05 2:59

>>69
No, linked structures are not inherently cache inefficient. There's nothing that says that linked structures somehow can't be close in memory.

Copying garbage collectors routinely move objects that refer to each other into a common shared region. Some fancy LISP collectors in the old and days of you're even removed the cdr pointer with the following pair itself (and so on).

In C you can easily put your list nodes in an array and allocate them from there. If the nodes get too spread out, move the nodes to one end (in sorted order).

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