Баловство с Haskell

module Test where
 
data Male = Man | Woman
 
male2 :: Male
male2 = Woman
 
instance Eq Male where
    (==) :: if Male -- Хм, что бы тут написато то... Male == Woman явно не прокатит.
 
fun :: Male -> String
fun a = if a == Woman
        then "Woman"
        else "Man"

module Test where
 
data Male = Man | Woman  deriving (Eq,Show)
 
fun :: Male -> String
fun Man = "Man"
fun Woman = "Woman"

instance Eq Male where
    Man   == Man   = True
    Woman == Woman = True
    _     == _     = False

data Male = Man | Woman
        deriving(Eq,Show)

 _     == _     = False

_  /= _ = Falsе

data Pair a b = Pair a b
u :: Either Int String -> Int -> Int
v :: Int -> Either Int String -> Int
w :: Int
 
f :: Pair (Either Int String) (Either Int String) -> Int
f (Pair x (Left y)) = u x y
f (Pair (Left x) y) = v x y
f _ _ = w

f _ _ … _ =

ghci> :t (==)
(==) :: Eq a => a -> a -> Bool
ghci> :i (==)
class Eq a where
  (==) :: a -> a -> Bool
  ...
    -- Defined in ‘GHC.Classes’
infix 4 ==

a -> a -> Bool

Eq a =>

ghci> x = _ 1
 
<interactive>:4:5: error:
    • Found hole: _ :: Integer -> t
      Where: ‘t’ is a rigid type variable bound by
               the inferred type of x :: t at <interactive>:4:1
    • In the expression: _
      In the expression: _ 1
      In an equation for ‘x’: x = _ 1
    • Relevant bindings include x :: t (bound at <interactive>:4:1)

GHCi, version 7.6.3: http://www.haskell.org/ghc/  :? for help
Loading package ghc-prim ... linking ... done.
Loading package integer-gmp ... linking ... done.
Loading package base ... linking ... done.
Prelude> :t _
 
<interactive>:1:1: Pattern syntax in expression context: _
Prelude> :t \x -> _
 
<interactive>:1:7: Pattern syntax in expression context: _
Prelude> x = _ 1
 
<interactive>:4:3: parse error on input `='
Prelude> let x = _ 1
 
<interactive>:5:9: Pattern syntax in expression context: _