Pushing rational numbers

Question

I have to make class NumberSet, which is a container for different kinds of numbers that can include only numbers that are not already in it.

class NumberSet
  include Enumerable

  def initialize
    @arr=[]
  end
  def each (&block)
    @arr.each do |member|
      block.call (member)
    end
  end
  def << number
    @arr<<number if @arr.include?(number) == false
  end
end

This code truncates the Rational numbers. For example, (22/7) should not equal (3/1).

mine=NumberSet.new
mine<<Rational(22/7)
# => [(3/1)]
mine<<3.0
# => nil

How can I fix this?

Is there a reason why you are not just using Ruby's `set` library? — Jörg W Mittag, Oct 21 '14 at 15:44

score 4 · Accepted Answer · answered Oct 21 '14 at 14:31

4

Your usage of Rational is wrong. Should be

mine << Rational(22, 7)

answered Oct 21 '14 at 14:31

Sergio Tulentsev

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Yep I discovered my mistake, silly me. Thank you for the answer. – user103220 Oct 21 '14 at 14:35
"22/7".to_r is another option. – steenslag Oct 21 '14 at 15:39

score 1 · Answer 2 · answered Oct 21 '14 at 14:40

1

With the newest Ruby, you should do:

mine << 22/7r

answered Oct 21 '14 at 14:40

sawa

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Should or could? There's a big difference :) – Sergio Tulentsev Oct 21 '14 at 14:40
@SergioTulentsev Should. Because that is less confusing and is concise. – sawa Oct 21 '14 at 14:41
2

I don't like this new syntax, but, fortunately, I never had to deal with rationals. So I don't really care. :) – Sergio Tulentsev Oct 21 '14 at 14:42
2

@Sergio, you are missing out on a lot in life by limiting yourself to irrational numbers. – Cary Swoveland Oct 21 '14 at 16:55

Pushing rational numbers

2 Answers2