HaskellQuiz's Answer

-- Quiz: http://blog.hackers-cafe.net/2010/06/haskell-quiz.html
-- Answer: http://blog.hackers-cafe.net/2010/07/haskell-quiz-answer.html

class Foo t where
    a :: t
    b :: t

instance Foo Int where
    a = 0
    b = 1

instance Foo Integer where
    a = 1
    b = 0

c :: Int
c = 0

d :: Integer
d = 0

main = do
  print $ a + c == 0 -- True.
  print $ a == c     -- True. a::Int == c::Int == 0.
  print $ c == 0     -- True. 
  print $ a + d == 1 -- True. 
  -- d is Integer. So the variable a is a::Integer, 
  -- which equals 1. "d == 1" is a misconception.
  print $ b + c == 1 -- True. b::Int is 1.
  print $ b + d == 0 -- True. However, b::Integer is 0.
  print $ b == d     -- True. Yes, both are 0::Integer;
  print $ d == 0     -- True. 
  -- Why I use "d == 0" not "b == 0"? 
  -- Because it causes "ambiguous type" error.
  -- You know, ":t 0" is "(Num t) => t". No way to determine
  -- it is Int or Integer.

  -- hints to make answer unique (perhaps...)
  print $ sum([a, b, c])  -- 1: It is [0, 1, 0]::[Int]
  print $ sum([a, b, d])  -- 1: It is [1, 0, 0]::[Integer]
  print $ [a, b, c] !! a  -- 0: (!!) take Int argument,
  print $ [a, b, c] !! b  -- 1: so it says (a::Int /= b::Int)
  print $ [a, b, c] !! c  -- 0
  -- Why I didn't show [a, b, c] !! d is, 
  -- it causes error because d is Integer.