我需要搞清楚一个编译错误这是真正推动我坚果一些帮助......Haskell的类型类和类型的家庭(续)
我有以下类型的类:
infixl 7 -->
class Selectable a s b where
type Res a s b :: *
(-->) :: (CNum n) => (Reference s a) -> (n,(a->b),(a->b->a)) -> Res a s b
我实例两次。第一次是这样一个魅力:
instance Selectable a s b where
type Res a s b = Reference s b
(-->) (Reference get set) (_,read,write) =
(Reference (\s ->
let (v,s') = get s
in (read v,s'))
(\s -> \x ->
let (v,s') = get s
v' = write v x
(_,s'') = set s' v'
in (x,s'')))
因为类型检查推断
(-->) :: Reference s a -> (n,a->b,a->b->a) -> Reference s b
与此签名与类签名匹配( - >),因为
Res a s b = Reference s b
现在我添加第二个实例并且所有内容都会中断:
instance (Recursive a, Rec a ~ reca) => Selectable a s (Method reca b c) where
type Res a s (Method reca b c) = b -> Reference s c
(-->) (Reference get set) (_,read,write) =
\(x :: b) ->
from_constant(Constant(\(s :: s)->
let (v,s') = get s :: (a,s)
m = read v
ry = m x :: Reference (reca) c
(y,v') = getter ry (cons v) :: (c,reca)
v'' = elim v'
(_,s'') = set s' v''
in (y,s''))) :: Reference s c
编译器抱怨
Couldn't match expected type `Res a s (Method reca b c)'
against inferred type `b -> Reference s c'
The lambda expression `\ (x :: b) -> ...' has one argument,
which does not match its type
In the expression:
\ (x :: b)
-> from_constant (Constant (\ (s :: s) -> let ... in ...)) ::
Reference s c
In the definition of `-->':
--> (Reference get set) (_, read, write)
= \ (x :: b)
-> from_constant (Constant (\ (s :: s) -> ...)) :: Reference s c
仔细阅读编译器告诉我,它已经推断出的类型( - >)正是如此:
(-->) :: Reference s a -> (n,a->(Method reca b c),a->(Method reca b c)->a) -> (b -> Reference s c)
这是正确的,因为
Res a s (Method reca b c) = b -> Reference s c
但为什么它不能匹配这两个定义?
对不起,不提供更succint和独立的例子,但在这种情况下,我想不出该怎么办呢?
你能给出一个完整的可运行示例吗?即从哪里引用?甚至包括'{ - #LANGUAGE TypeFamilies# - }'。这可以帮助别人帮助你。 – yairchu 2010-04-08 09:34:36
这是一个很大的...这只是一个更大的项目,我不可能完整发布:( – 2010-04-08 19:37:35