2017-08-13 99 views
0

最近我一直在探索Idris中的依赖类型。然而,我克服了一个非常烦人的问题,这是在伊德里斯,我应该用类型签名开始我的程序。所以问题是,如何在Idris中编写简洁的类型签名?如何在Idris中编写Vect的正确类型签名?

例如,

get_member : (store : Vect n String) -> (idx : List (Fin n)) -> (m : Nat ** Vect m (Nat, String)) 
get_member store idx = Vect.mapMaybe maybe_member (Vect.fromList idx) 
    where 
    maybe_member : (x : Fin n) -> Maybe (Nat, String) 
-- The below line should be type corrected 
-- maybe_member x = Just (Data.Fin.finToNat x, Vect.index x store) 

如果我评论的最后一行,编译将键入检查上述功能, 但如果我做的最后行表达,编译会报错。

When checking right hand side of VecSort.get_member, maybe_member with expected type 
     Maybe (Nat, String) 
When checking an application of function Data.Vect.index: 
     Type mismatch between 
       Vect n1 String (Type of store) 
     and 
       Vect n String (Expected type) 

     Specifically: 
       Type mismatch between 
         n1 
       and 
         n 

但我这样做是为lambda功能,

get_member : (store : Vect n String) -> (idx : List (Fin n)) -> (m : Nat ** Vect m (Nat, String)) 
get_member store idx = Vect.mapMaybe (\x => Just (Data.Fin.finToNat x, Vect.index x store)) (Vect.fromList idx) 

这将是类型检查为好。

所以问题是,我应该如何定义类型签名中正确长度的Vect类型?

回答

0

我不知道如果我的解释是正确的,但以下typechecks:

get_member : (store : Vect n String) -> (idx : List (Fin n)) -> (m : Nat ** Vect m (Nat, String)) 
get_member store idx {n} = Vect.mapMaybe (maybe_member) (Vect.fromList idx) 
    where 
     maybe_member : (x : Fin n) -> Maybe (Nat, String) 
     maybe_member x = Just (Data.Fin.finToNat x, Vect.index x store) 

不同的是,你进入你的范围的隐含参数n。这{n}x: Fin x都提到相同的n。如果不在您的范围内拉取隐含的n,idris不能假定两个n确实是相同的,并且它会抱怨错误消息,它不知道n1n是否相同。