Haskell:我的缩进错误是什么

Question

我通过以下代码得到了我认为的缩进错误

152     -- find the first one who can refute the scenario
153     let cs = map ($scenario) [ Suspect . getWho, Room . getWhere, Weapon . getHow ]
154     let (qs,rss) = break (not . null . intersect cs . hand) ps
155     unless (null rss) $ do
156       let refuter:ss= rss
157       let valid = intersect cs $ hand refuter
158 
159       (refuter, shown) <- if cheated refuter
160                             -- if the refuter is a cheater, just choose for him
161                             then return (refuter, head valid)
162                             else do
163                               (refuter, shown) <- runStateT $ reveal scenario
164                               -- make sure they don't cheat
165                               return if shown `elem` valid
166                                         then (refuter, shown)
167                                         else (refuter { cheated = True }, head valid)
168       putTail (qs ++ refuter:ss)
169     return ()

我得到的错误是

Game.hs:159:6:
    The last statement in a 'do' construct must be an expression

当我删除或注释掉第159-167行时,错误消失了.

我究竟做错了什么？

Answer 1

你需要return在第165行的值中的if-then-else周围加上括号:

return (if shown `elem` valid
           then (refuter, shown)
           else (refuter { cheated = True }, head valid))

Haskell语法不允许将纯if-then-else 作为函数参数,它必须放在括号中以便在该位置使用.(它是一个lexp,而函数参数必须是一个aexp.)

生成的错误消息对查找该问题不是很有帮助.