haskell

来源： https://stackoverflow.com/questions/62764261/functional-tail-recursive-way-to-generate-all-possible-combinations-from-a-dict

来源： https://stackoverflow.com/questions/62764261/functional-tail-recursive-way-to-generate-all-possible-combinations-from-a-dict

来源： https://stackoverflow.com/questions/62537779/reinstalling-the-same-version-of-a-package-with-cabal-new-install

来源： https://stackoverflow.com/questions/63191868/array-aggregation-across-a-link-table-in-opaleye

来源： https://stackoverflow.com/questions/63191868/array-aggregation-across-a-link-table-in-opaleye

来源： https://stackoverflow.com/questions/36655974/what-is-the-difference-between-xor-and-or-in-haskell

来源： https://stackoverflow.com/questions/36655974/what-is-the-difference-between-xor-and-or-in-haskell

问题 I need to make function returns all possible branches from a tree with this form: data Tree a = EmptyT | NodeT a ( Tree a ) ( Tree a ) deriving (Show) everyBranch :: Tree a -> [[a]] I'm not sure how to approach this... xD I'm still a newbie in Haskell. Let's say that I have: 1 / \ 2 3 /\ / \ 4 5 7 8 I want to get: [[1,2,4], [1,2,5], [1,3,8], [1,3,7]] 回答1: We'll use a recursive approach. Let's start with a rough skeleton: everyBranch :: Tree a -> [[a]] everyBranch EmptyT = _something

问题 I'm kind of stuck with an assignement concerning my exams. I want to find out the types of those two functions by applying the unifying algorithm by hand: map map (\x -> x >>= (\y -> y)) Could someone point me to the right direction? The only ressource I could find until now was the wikipedia entry which is not really aiding me because of the high level of abstraction. Greetings and thank you. 回答1: Let's just do the first. map :: (a -> b) -> [a] -> [b] Now we can write it again with two

问题 I'm kind of stuck with an assignement concerning my exams. I want to find out the types of those two functions by applying the unifying algorithm by hand: map map (\x -> x >>= (\y -> y)) Could someone point me to the right direction? The only ressource I could find until now was the wikipedia entry which is not really aiding me because of the high level of abstraction. Greetings and thank you. 回答1: Let's just do the first. map :: (a -> b) -> [a] -> [b] Now we can write it again with two