我正在玩idris,并试图实现可扩展的记录。
主要目标是保证记录的密钥是唯一的。
例如
("Year" := 1998) +: rnil
是合法的
("Title" := "test") +: ("Year" := 1998) +: rnil
也有效,但
("Year" := "test") +: ("Year" := 1998) +: rnil
无法编译。
我提出了以下实现,编译得很好:
{-
See: http://lpaste.net/104020
and https://github.com/gonzaw/extensible-records
-}
module Record
import Data.List
%default total
data HList : List Type -> Type where
Nil : HList []
(::) : a -> HList xs -> HList (a :: xs)
infix 5 :=
data Field : lbl -> Type -> Type where
(:=) : (label : lbl) ->
(value : b) -> Field label b
labelsOf : List (lbl, Type) -> List lbl
labelsOf [] = []
labelsOf ((label, _) :: xs) = label :: labelsOf xs
toFields : List (lbl, Type) -> List Type
toFields [] = []
toFields ((l, t) :: xs) = (Field l t) :: toFields xs
data IsSet : List t -> Type where
IsSetNil : IsSet []
IsSetCons : Not (Elem x xs) -> IsSet xs -> IsSet (x :: xs)
data Record : List (lty, Type) -> Type where
MkRecord : IsSet (labelsOf ts) -> HList (toFields ts) ->
Record ts
infixr 6 +:
rnil : Record []
rnil = MkRecord IsSetNil []
prepend : { label : lbl,
xs : List (lbl, Type),
prf : Not (Elem label (labelsOf xs))
} ->
Field label t ->
Record xs ->
Record ((label, t) :: xs)
prepend {prf} f (MkRecord isSet fs) = MkRecord (IsSetCons prf isSet) (f :: fs)
data IsNo : Dec prop -> Type where
ItIsNo : IsNo (No y)
(+:) : DecEq lbl =>
{ label : lbl, xs : List (lbl, Type) } ->
Field label t ->
Record xs ->
{ auto isno : IsNo (isElem label $ labelsOf xs) } ->
Record ((label, t) :: xs)
(+:) {label} {xs} f r with (isElem label $ labelsOf xs)
(+:) { isno = ItIsNo } _ _ | (Yes _) impossible
(+:) f r | (No no) = prepend {prf = no} f r
有趣的是
{ auto isno : IsNo (isElem label $ labelsOf xs) } ->
这个想法是,如果密钥是唯一的,编译器将琐事地找到IsNo
的实例,而如果密钥不是唯一的,那么它就不会被编译。
这适用于那些示例
("Year" := 1998) +: rnil -- Compiles fine
("Year" := "test") +: ("Year" := 1998) +: rnil -- fails to compile as expected
但是
("Title" := "test") +: ("Year" := 1998) +: rnil
无法使用以下错误进行编译:
(input):Type mismatch between
("Title" = "Year") -> "Title" = "Year"
and
("Title" = "Year") -> Void
我必须承认这个错误令我感到困惑。谁能解释一下这里发生了什么?
答案 0 :(得分:3)
您似乎是第一个在愤怒中使用DecEq
String
实例的人,因此,您是第一个发现我们在这里为基元构建证明术语的方式是错的。对于那个很抱歉。好消息是它很容易修复(我刚刚在你的例子中试过它并且它很好),坏消息是你在推动修复后需要git头。
无论如何,我们已经推迟了新的发布。我会在本周末尝试这样做。你的代码对我来说当然很好!