为什么这两个Haskell"展开"功能不同？

Question

我正在学习Haskell,现在我正在使用Maybe Class进行练习.我必须创建一个函数,它将f("Maybe function")重复应用于(及其后续结果),直到f a返回Nothing.例如f a0 = Just a1,f a1 = Just a2,...,f an = Nothing.然后

unfold f a0 = [a0,a1,...,an]

我试过这样做而且我得到了:

unfold :: (a- > Maybe a) -> a -> [a]
unfold f a = case f a of
                 Just n -> n: unfold f a
                 Nothing -> []

问题是解决方案是:

unfold' :: ( a -> Maybe a) -> a -> [a]
unfold' f a = a : rest ( f a )
     where rest Nothing = []
           rest ( Just x ) = unfold' f x

而我的程序不像解决方案那样工作.也许我使用了错误的"案例",但我不确定.

Answer 1

您的使用case很好,但请查看列表中的新值以及解决方案的位置.

testFunc = const Nothing

unfold  testFunc 1 == []  -- your version prepends only if f a isn't Nothing
unfold' testFunc 1 == [1] -- the solution _always_ prepends the current value

此外,您始终使用相同的值.

unfold :: (a -> Maybe a) ->a -> [a]
unfold f a = a : case f a of -- cons before the case
    Just n  -> unfold f n    -- use n as parameter for f
    Nothing -> []