The Little Schemer (ver. Haskell), Scheme 手習い （Haskell 版）

gistfile1.hs

-- Scheme 手習い 5章 p.83
-- The Little Schemer Section 5
rember_star :: (Eq a, Eq b) => a -> [b] -> [b]
rember_star _ [] = []
rember_star e (x:xs) =
  case x of
    ys@(_:_)   -> (rember_star e ys) : (rember_star e xs)
    z | e == z -> rember_star e xs
    _          -> x : rember_star e xs

-- haskeller.hs:7:34:
--     Could not deduce (a ~ [t0])
-- haskeller.hs:8:14:
--     Could not deduce (b ~ [t0])


{-
このままの構造だと実装は無理な気がする。どうすべきか。
I think that rember_star cannot be implemented with this structure. How should I implement it?
-}

{-
; quote The Little Schemer
(define rember*
  (lambda (a l)
    (cond
      ((null? l) (quote ())
      ((atom? (car l))
        (cond
          ((eq? (car l) a) (rember* a (cdr l)))
          (else (cons (car l) (rember* a (cdr l))))))
      (else (cons (rember* a (car l)) (rember* a (cdr l)))))))
-}

少なくともリストの要素の型が全て同じじゃないと Haskell で実装できないよな。