使依赖参数在 Functor 定义中可用
Make dependent parameter available in Functor definition
我有以下 Vect
的包装器
data Foo : (r : Nat) -> (t: Type) -> Type where
MkFoo : Vect r t -> Foo r t
我想为 Foo
实现 Functor
,并在实现中使用 r
的值,但我无法编译它。
Functor (Foo r) where
map f (MkFoo v) = MkFoo (ff v) where
ff : {r' : Nat} -> Vect r' a -> Vect r' b
给予
Error: While processing right hand side of map. r is not accessible in this
context.
Foo.idr:61:28--61:32
|
61 | map f (MkFoo v) = MkFoo (ff v) where
| ^^^^
这在 Idris 1 中有效,但我不知道如何将它移植到 Idris2。我试过 r
不被删除,
Functor ({r : _} -> Foo r) where
map f (MkFoo v) = MkFoo (ff v) where
ff : {r' : Nat} -> Vect r' a -> Vect r' b
但我明白了
Error: While processing type
of Functor implementation at Foo.idr:60:1--62:46. When
unifying Type -> Type and Type.
Mismatch between: Type -> Type and Type.
Foo.idr:60:21--60:26
|
60 | Functor ({r : _} -> Foo r) where
| ^^^^^
尝试:
{r : _} -> Functor (Foo r) where
map f (MkFoo v) = MkFoo (ff v) where
ff : {r' : Nat} -> Vect r' a -> Vect r' b
我有以下 Vect
data Foo : (r : Nat) -> (t: Type) -> Type where
MkFoo : Vect r t -> Foo r t
我想为 Foo
实现 Functor
,并在实现中使用 r
的值,但我无法编译它。
Functor (Foo r) where
map f (MkFoo v) = MkFoo (ff v) where
ff : {r' : Nat} -> Vect r' a -> Vect r' b
给予
Error: While processing right hand side of map. r is not accessible in this
context.
Foo.idr:61:28--61:32
|
61 | map f (MkFoo v) = MkFoo (ff v) where
| ^^^^
这在 Idris 1 中有效,但我不知道如何将它移植到 Idris2。我试过 r
不被删除,
Functor ({r : _} -> Foo r) where
map f (MkFoo v) = MkFoo (ff v) where
ff : {r' : Nat} -> Vect r' a -> Vect r' b
但我明白了
Error: While processing type
of Functor implementation at Foo.idr:60:1--62:46. When
unifying Type -> Type and Type.
Mismatch between: Type -> Type and Type.
Foo.idr:60:21--60:26
|
60 | Functor ({r : _} -> Foo r) where
| ^^^^^
尝试:
{r : _} -> Functor (Foo r) where
map f (MkFoo v) = MkFoo (ff v) where
ff : {r' : Nat} -> Vect r' a -> Vect r' b