haskell - 如何获取字符串 'aa' 、 ab' 到 'yz' 、 'zz' ？

Question

如何获取列表中从“aa”到“zz”的字符串？我知道这是显而易见的，但不知道解决此类问题的正确习惯用法。只需用具体的例子来展示这个想法，剩下的我会弄清楚。谢谢。

尝试过

(++) <$> ['a'..'z'] <*> ['a'..'z']

但它无法编译。

Answer 1

所有这些都可以满足您的要求(记住String = [Char]):

Control.Monad.replicateM 2 ['a'..'z'] -- Cleanest; generalizes to all Applicatives
sequence $ replicate 2 ['a'..'z'] -- replicateM n = sequenceA . replicate n
-- sequence = sequenceA for monads
-- sequence means cartesian product for the [] monad

[[x, y] | x <- ['a'..'z'], y <- ['a'..'z']] -- Easiest for beginners
do x <- ['a'..'z']
   y <- ['a'..'z']
   return [x, y] -- For when you forget list comprehensions exist/need another monad
['a'..'z'] >>= \x -> ['a'..'z'] >>= \y -> return [x, y] -- Desugaring of do
-- Also, return x = [x] in this case
concatMap (\x -> map (\y -> [x, y]) ['a'..'z']) ['a'..'z'] -- Desugaring of comprehension
-- List comprehensions have similar syntax to do-notation and mean about the same,
-- but they desugar differently

(\x y -> [x, y]) <$> ['a'..'z'] <*> ['a'..'z'] -- When you're being too clever
(. return) . (:) <$> ['a'..'z'] <*> ['a'..'z'] -- Same as ^ but pointfree

原因

(++) <$> ['a'..'z'] <*> ['a'..'z']

不起作用是因为你需要 (++)::Char -> Char -> [Char]，但你只有 (++)::[Char] -> [字符] -> [字符]。您可以在 (++) 的参数顶部添加 returns ，以将 Char 放入单例列表中并让事情正常工作:

(. return) . (++) . return <$> ['a'..'z'] <*> ['a'..'z']