Confused about Monad output

[Dean Herington]

You have an extra `show` in the `Show` instance for `Out`.
Change `show x` to `x` there.

On Wed, 30 Jan 2002, Shawn P. Garbett wrote:

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> I've been fiddling with the example in section 10.2 of Bird's book, 
> _Introduction_to_Functional_Programming_. I can't seem to make sense of why 
> the output appears as it does. It's as if the final evaluation through Show 
> doesn't take place, is there some critical piece of syntax missing?
> 
> I expect:
> 
> Main> evalOut(answer)
> term: Con 1972, yields 1972
> term: Con 2, yields 2
> term: Div (Con 1972) (Con 2), yields 986
> term: Con 23, yields 23
> term: Div(Div(Con 1972)(Con 2))(Con 23), yields 42
> value: 42
> 
> but I get: 
> 
> Main> evalOut(answer)
> ("term: Con 1972, yields 1972\nterm: Con 2, yields 2\nterm: Div (Con 1972) 
> (Con 2), yields 986\nterm: Con 23, yields 23\nterm: Div (Div (Con 1972) (Con 
> 2)) (Con 23), yields 42\n",42)
> (1149 reductions, 3134 cells)
> Main> 
> 
> 
> Here's the program:
> 
> data Term = Con Int | Div Term Term
>             deriving(Show)
> 
> - -- 10.2.5 Monadic evaluator 
> eval            :: Monad m => Term -> m Int
> eval (Con x)    =  return x
> eval (Div t u)  =  do x <- eval t
>                       y <- eval u
>                       return (x `div` y)
> 
> - -- Examples to try
> 
> answer, wrong :: Term
> answer = Div ( Div ( Con 1972 ) (Con 2)) (Con 23)
> wrong  = Div (Con 2) (Div (Con 1)(Con 0))
> 
> - -- Output monad
> newtype Out a = MkOut (String, a)
> 
> instance Monad Out where 
>   return x  =  MkOut ("", x)
>   p >>= q   =  MkOut (ox ++ oy, y)
>                where MkOut (ox, x) = p
>                      MkOut (oy, y) = q x
> 
> instance Show a => Show (Out a) where
>   show (MkOut (x,y)) = show x ++ "value: " ++ show y
> 
> - -- operation to generate output
> 
> 
> out     :: String -> Out()
> out ox  =  MkOut (ox, ())
> 
> 
> line      :: Term -> Int -> String
> line t x  =  "term: " ++ show t ++ ", yields " ++ show x ++ "\n"
> 
> evalOut          :: Term -> Out Int
> evalOut (Con x)  =  do out (line(Con x) x)
>                        return x
> evalOut (Div t u) = do x <- evalOut t
>                        y <- evalOut u
>                        out(line(Div t u) (x `div` y))
>                        return (x `div` y)
> 
> 
> Shawn