最好的写法是-> ..->[a] haskell 中的递归函数
所以我一直遇到这个小问题，
func :: a -> b -> [a] -- # or basically any a-> ...-> [a] where ... is any types ->
func x y = func' [x] y -- '# as long as they are used to generate a list of [a] from x
func' :: [a] -> b -> [a]
func' = undefined -- # situation dependant generates a list from
-- # each element and returns it as one long list
我应该保持这样吗？
我应该使用隐藏在何处的 func' 吗？
我应该只使用 [a] -> b-> [a] 版本并将传递[变量] 的责任留给被调用者？
我可能很需要组合这些函数，并且可能想打乱顺序，所以我倾向于选项 3。您觉得怎么样？
So I keep having this small problem where I have something like
func :: a -> b -> [a] -- # or basically any a-> ...-> [a] where ... is any types ->
func x y = func' [x] y -- '# as long as they are used to generate a list of [a] from x
func' :: [a] -> b -> [a]
func' = undefined -- # situation dependant generates a list from
-- # each element and returns it as one long list
should I keep it like this?
should I use func' hidden by a where?
should I only use the [a] -> b -> [a] version and leave the responsibility of passing [variable] to the callee?
I might well need to compose these functions and might want to mess around with the order so I'm leaning towards option 3. What do you think?
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看起来您正在尝试重新发明 concatMap：
因此“map”位将输入列表的每个元素应用“f”。 “f”接受单个“a”并返回“[b]”。然后将这些单独的列表连接成一个列表。
It looks like you are trying to reinvent concatMap:
So the "map" bit takes each element of the input list applies "f" to it. "f" takes a single "a" and returns a "[b]". These individual lists are then concatenated into a single list.
正如 Paul 指出的，
func'可以替换为concatMap和func。func本身让我想起了unfoldrfrom Data.List：用于从
生成。a列表b.顺便说一句，
func和func'对于此类函数来说是不幸的名称。As Paul noted,
func'can be replaced withconcatMapandfunc.And
funcitself reminds me ofunfoldrfrom Data.List:It is used to generate a list of
afromb.By the way,
funcandfunc'are unfortunate names for such functions.