From: "Jan E." <lists@...>
Date: 2012-09-28T18:08:52+09:00
Subject: Re: calcaulation with unknown numbers of numbers and options fail

Hi,

Roelof Wobben wrote in post #1077862:
> I can declare it as this  def calculate (*arguments)
> So i will be one hash but then I have to figure out how to seperate the
> options part and the numbers parts.

Then why do you do it? Just set the parameter list according to the 
method call:

def calculate *numbers, options
  ...
end

I'm also quite sure that you're not supposed to subtract the numbers 
from 0 but rather from each other. So calculate(1, 2, {}) should yield 1 
- 2 = -1 and not 0 - 1 - 2 = -3.

In this case you must omit the "0" parameter for "inject". And you 
should use the short form:


def calculate *numbers, options
  operator = options[:add] ? :+ : :-
  numbers.reduce operator
end

puts calculate 3, 5, add: true
puts calculate 3, 5, add: false
puts calculate(3, 5, {})

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