[英]How to correctly return nested types in haskell?
I want to return empty lists if one of the parameters is Nothing, but I can't如果参数之一为 Nothing,我想返回空列表,但我不能
data UtilsDiff = UtilsDiff {syntaxBlockOptions :: [String], optionsBlockOptions :: [String]}
deriving Show
data UtilsOrDiff = UtilsOrDiff Utils | Diff UtilsDiff
deriving Show
useBlockExpr :: Parser UtilsOrDiff
useBlockExpr = do
u1 <- syntaxExpr
opts <- optionsBlockExpr
_ <- absorbToStopWord "Description:"
utilDesc <- many anyChar
case options u1 of
Nothing -> Diff [] [] -- <- this line
Just val-> do
...
Apparently I haven't figured out exactly how to return the Diff type显然我还没有弄清楚如何返回 Diff 类型
warning: [-Wdeferred-type-errors]
• Couldn't match expected type ‘[a0] -> Text.Parsec.Prim.ParsecT String () Data.Functor.Identity.Identity UtilsOrDiff’ with actual type ‘UtilsOrDiff’
• The function ‘Diff’ is applied to two value arguments,but its type ‘UtilsDiff -> UtilsOrDiff’ has only one
In the expression: Diff [] [] In a case alternative: Nothing -> Diff [] []
Diff
has type UtilsDiff -> UtilsOrDiff
, not [a] -> [a] -> UtilsDiff
. Diff
的类型为UtilsDiff -> UtilsOrDiff
,而不是[a] -> [a] -> UtilsDiff
。 You need to create the UtilsDiff
value first, then use that to create the Diff
value.您需要先创建
UtilsDiff
值,然后使用它来创建Diff
值。
Nothing -> return $ Diff (UtilsDiff [] [])
As the Just
branch presumably evaluates to a Parser UtilsOrDiff
value, so must the Nothing
branch, hence the call to return
.由于
Just
分支可能评估为Parser UtilsOrDiff
值,因此Nothing
分支也必须如此,因此调用return
。
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