如何获得字符串'aa'，ab'到'yz'，'zz'？

Question

如何从列表中的'aa'到'zz'获取字符串？ 我知道这很明显，但不知道解决这类问题的恰当习惯。只要用具体的例子来展示这个想法，我就会找出其余的。 谢谢。

试图

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

但它不会编译。

Answer 1

所有这些做什么，你想要的（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'] 

不工作是因为你需要(++) :: Char -> Char -> [Char]，但你只有(++) :: [Char] -> [Char] -> [Char]。您可以在returns S于的参数上面扔(++)把Char s转换单列表，并把事情的工作：

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