Class equality with coercion - ruby

Question

I created this class based on predefined tests
class Roman
include Comparable
ROMAN_NUMBERS = { 1000 => "m", 900 => "cm", 500 => "d", 400 => "cd",
100 => "c", 90 => "xc", 50 => "l", 40 => "xl", 10 => "x", 9 => "ix",
5 => "v", 4 => "iv", 1 => "i" }
def initialize(number)
#number = number
end
def coerce(other)
if Integer === other
[ other, #number ]
else
[ Float(other), Float(#number) ]
end
end
def <=>(n)
#number <=> n
end
def -(n)
Roman.new(#number - n)
end
def to_s
roman = ''
ROMAN_NUMBERS.each do |value, letter|
roman << letter * (#number / value)
#number = #number % value
end
roman
end
end
but I'm still failing this one
it "should support substraction" do
(Roman.new(5) - Roman.new(3)).should == 2
(Roman.new(4) - Roman.new(1)).should == 3
(Roman.new(4) - 2).should == 2
res = 6 - Roman.new(1)
res.should == 5
(res.kind_of? Roman).should be true
end
The problem is, that res is "kind of" Fixnum. How's that possible? How to make it pass the test?

Answer 1

The problem is that 6 - Roman.new(1) is the same as 6.-(Roman.new(1)) i.e. it is calling the built in subtraction method of Fixnum. Since Fixnum doesn't know about your Roman class, this calls your coerce method.
But you defined coerce to return a Fixnum in this case! The solution is to coerce the other operand, rather than self.
def coerce other
[self.class.new(other), self]
end
But if you try this with your code, you'll get a stack error! That's because it's now coercing both operands into instances of Roman and it still doesn't know how to subtract them. You should redefine your subtraction method to handle those cases.
attr_reader :number
def - n
return Roman.new(#number - n.number) if n.is_a? Roman
Roman.new(#number - n)
end