I just found an old post from @evanilukhin showing interview task - Simple word calculator. I was reading it on my mobile, so first I read his post, and decided to not look at the answer code (anyway it's impossible to understand on small android display).
Task is next: make this code work:
one.plus.two.equal # => 3 one.minus.three.equal # => -2 While you're on interview, it can be harder than it should be. Few minutes later electricity came back (some works during this morning... awful), so I decided to try to solve it (in his post I read it's good for every developer to check his skills... and it's true). You have a link to his post if you want to read it and see his results (probably better than mine, but hey, 10 minutes on android 5, and it works... :P).
So what do you think, are you able to make this code work without reading the rest? Here's my solution (first one, fast one):
and about 10 minutes later I decided to make this more DRY, and to allow it to use 'one' chained more times (like original article code work)My Answer
class One def initialize @num = 1 end def plus @opt = :add return self end def minus @opt = :rmw return self end def two case @opt when :add then @num += 2 when :rmw then @num -= 2 end return self end def three case @opt when :add then @num += 3 when :rmw then @num -= 3 end return self end def equal num = @num @num = 1 return num end end one = One.new one.plus.two.equal => 3 one.minus.three.equal => -2
class One def initialize @num = 1 end def one use_number 1 end def two use_number 2 end def three use_number 3 end def plus; set_opt(:add) end def minus; set_opt(:rmw) end def equal num = @num and @num = 1 return num end private def set_opt(opt) @opt = opt; return self end def use_number(number) case @opt when :add then @num += number.to_i when :rmw then @num -= number.to_i else @num end @opt = nil return self end end one = One.new one.plus.two.equal # => 3 one.plus.two.minus.three.plus.one.equal => 1
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