Question

如何从＆＃39; aa＆＃39;中获取字符串。直至＆＃39; zz＆＃39;在列表中？我知道这很明显，但不知道解决这类问题的正确习惯。只要用具体的例子来表达这个想法，我就会弄清楚其余部分。感谢。

尝试

std::remove_copy(source.begin(), source.end(), std::back_inserter(dest), value);

但它没有编译。

Answer 1

所有这些都做你想要的（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']

不起作用是因为您需要(++) :: Char -> Char -> [Char]，但您只有(++) :: [Char] -> [Char] -> [Char]。您可以在returns的参数之上放入(++)以将Char放入单例列表中并使其起作用：

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

如何获得字符串＆＃39; aa＆＃39;，ab＆＃39;直至＆＃39; yz＆＃39;，＆＃39; zz＆＃39;？

1 个答案: