Type kind error when creating instance of Functor typeclass
I'm trying to implement Functor typeclass instance for a very trivial type Foo:
data Foo a = Foo a
instance functorFoo :: Functor (Foo a) where
map fn (Foo a) = Foo (fn a)
Purescript gives me not-so-helpful error message:
Could not match kind
Type -> Type
with kind
Type
What does it mean? I'm not yet really familiar with the Kind-system.
typeclass functor purescript
asked Nov 22 '18 at 12:14
Rene SaarsooRene Saarsoo
9,05384567
add a comment |
I'm trying to implement Functor typeclass instance for a very trivial type Foo:
data Foo a = Foo a
instance functorFoo :: Functor (Foo a) where
map fn (Foo a) = Foo (fn a)
Purescript gives me not-so-helpful error message:
Could not match kind
Type -> Type
with kind
Type
What does it mean? I'm not yet really familiar with the Kind-system.
typeclass functor purescript
asked Nov 22 '18 at 12:14
Rene SaarsooRene Saarsoo
9,05384567
add a comment |
I'm trying to implement Functor typeclass instance for a very trivial type Foo:
data Foo a = Foo a
instance functorFoo :: Functor (Foo a) where
map fn (Foo a) = Foo (fn a)
Purescript gives me not-so-helpful error message:
Could not match kind
Type -> Type
with kind
Type
What does it mean? I'm not yet really familiar with the Kind-system.
typeclass functor purescript
asked Nov 22 '18 at 12:14
Rene SaarsooRene Saarsoo
9,05384567
I'm trying to implement Functor typeclass instance for a very trivial type Foo:
data Foo a = Foo a
instance functorFoo :: Functor (Foo a) where
map fn (Foo a) = Foo (fn a)
Purescript gives me not-so-helpful error message:
Could not match kind
Type -> Type
with kind
Type
What does it mean? I'm not yet really familiar with the Kind-system.
typeclass functor purescript
typeclass functor purescript
asked Nov 22 '18 at 12:14
Rene SaarsooRene Saarsoo
9,05384567
asked Nov 22 '18 at 12:14
Rene SaarsooRene Saarsoo
9,05384567
asked Nov 22 '18 at 12:14
Rene SaarsooRene Saarsoo
9,05384567
asked Nov 22 '18 at 12:14
Rene SaarsooRene Saarsoo
9,05384567
asked Nov 22 '18 at 12:14
Rene SaarsooRene Saarsoo
9,05384567
9,05384567
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
But I'd still like to find out why does the first version not work. Like how does the Functor typeclass differ from e.g. Semigroup where the first version worked just fine.
Let's look at the definition of the Functor type class:
class Functor f where
map :: forall a b. (a -> b) -> f a -> f b
It contains f a and f b types in the type signature of map which clearly indicates that f is of kind Type -> Type (it "carries" another type). On the other hand Semigroup clearly relates to plain types of kind Type:
class Semigroup a where
append :: a -> a -> a
So Semigroup is a class which can be defined for plain types like String, Int, List a, Map k v or even a function a -> b (which can also be written as (->) a b) but not type constructors which require another type like List Map k (->) a (I have to use this notation here).
On the other hand Functor class requires type constructors so you can have instances like Functor List, Functor (Map k) or Functor ((->) a).
answered Nov 22 '18 at 22:36
paluhpaluh
1,5711511
Thanks for a through explanation. Much appreciated!
– Rene Saarsoo
Nov 23 '18 at 7:43
add a comment |
I found a solution. Instead of defining it as:
instance functorFoo :: Functor (Foo a) where
I need to define it as:
instance functorFoo :: Functor Foo where
But I'd still like to find out why does the first version not work. Like how does the Functor typeclass differ from e.g. Semigroup where the first version worked just fine.
answered Nov 22 '18 at 12:22
Rene SaarsooRene Saarsoo
9,05384567
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
But I'd still like to find out why does the first version not work. Like how does the Functor typeclass differ from e.g. Semigroup where the first version worked just fine.
Let's look at the definition of the Functor type class:
class Functor f where
map :: forall a b. (a -> b) -> f a -> f b
It contains f a and f b types in the type signature of map which clearly indicates that f is of kind Type -> Type (it "carries" another type). On the other hand Semigroup clearly relates to plain types of kind Type:
class Semigroup a where
append :: a -> a -> a
So Semigroup is a class which can be defined for plain types like String, Int, List a, Map k v or even a function a -> b (which can also be written as (->) a b) but not type constructors which require another type like List Map k (->) a (I have to use this notation here).
On the other hand Functor class requires type constructors so you can have instances like Functor List, Functor (Map k) or Functor ((->) a).
answered Nov 22 '18 at 22:36
paluhpaluh
1,5711511
Thanks for a through explanation. Much appreciated!
– Rene Saarsoo
Nov 23 '18 at 7:43
add a comment |
But I'd still like to find out why does the first version not work. Like how does the Functor typeclass differ from e.g. Semigroup where the first version worked just fine.
Let's look at the definition of the Functor type class:
class Functor f where
map :: forall a b. (a -> b) -> f a -> f b
It contains f a and f b types in the type signature of map which clearly indicates that f is of kind Type -> Type (it "carries" another type). On the other hand Semigroup clearly relates to plain types of kind Type:
class Semigroup a where
append :: a -> a -> a
So Semigroup is a class which can be defined for plain types like String, Int, List a, Map k v or even a function a -> b (which can also be written as (->) a b) but not type constructors which require another type like List Map k (->) a (I have to use this notation here).
On the other hand Functor class requires type constructors so you can have instances like Functor List, Functor (Map k) or Functor ((->) a).
answered Nov 22 '18 at 22:36
paluhpaluh
1,5711511
Thanks for a through explanation. Much appreciated!
– Rene Saarsoo
Nov 23 '18 at 7:43
add a comment |
But I'd still like to find out why does the first version not work. Like how does the Functor typeclass differ from e.g. Semigroup where the first version worked just fine.
Let's look at the definition of the Functor type class:
class Functor f where
map :: forall a b. (a -> b) -> f a -> f b
It contains f a and f b types in the type signature of map which clearly indicates that f is of kind Type -> Type (it "carries" another type). On the other hand Semigroup clearly relates to plain types of kind Type:
class Semigroup a where
append :: a -> a -> a
So Semigroup is a class which can be defined for plain types like String, Int, List a, Map k v or even a function a -> b (which can also be written as (->) a b) but not type constructors which require another type like List Map k (->) a (I have to use this notation here).
On the other hand Functor class requires type constructors so you can have instances like Functor List, Functor (Map k) or Functor ((->) a).
answered Nov 22 '18 at 22:36
paluhpaluh
1,5711511
But I'd still like to find out why does the first version not work. Like how does the Functor typeclass differ from e.g. Semigroup where the first version worked just fine.
Let's look at the definition of the Functor type class:
class Functor f where
map :: forall a b. (a -> b) -> f a -> f b
It contains f a and f b types in the type signature of map which clearly indicates that f is of kind Type -> Type (it "carries" another type). On the other hand Semigroup clearly relates to plain types of kind Type:
class Semigroup a where
append :: a -> a -> a
So Semigroup is a class which can be defined for plain types like String, Int, List a, Map k v or even a function a -> b (which can also be written as (->) a b) but not type constructors which require another type like List Map k (->) a (I have to use this notation here).
On the other hand Functor class requires type constructors so you can have instances like Functor List, Functor (Map k) or Functor ((->) a).
answered Nov 22 '18 at 22:36
paluhpaluh
1,5711511
edited Nov 24 '18 at 23:03
answered Nov 22 '18 at 22:36
paluhpaluh
1,5711511
answered Nov 22 '18 at 22:36
paluhpaluh
1,5711511
answered Nov 22 '18 at 22:36
paluhpaluh
1,5711511
1,5711511
Thanks for a through explanation. Much appreciated!
– Rene Saarsoo
Nov 23 '18 at 7:43
add a comment |
Thanks for a through explanation. Much appreciated!
– Rene Saarsoo
Nov 23 '18 at 7:43
Thanks for a through explanation. Much appreciated!
– Rene Saarsoo
Nov 23 '18 at 7:43
Thanks for a through explanation. Much appreciated!
– Rene Saarsoo
Nov 23 '18 at 7:43
add a comment |
I found a solution. Instead of defining it as:
instance functorFoo :: Functor (Foo a) where
I need to define it as:
instance functorFoo :: Functor Foo where
But I'd still like to find out why does the first version not work. Like how does the Functor typeclass differ from e.g. Semigroup where the first version worked just fine.
answered Nov 22 '18 at 12:22
Rene SaarsooRene Saarsoo
9,05384567
add a comment |
I found a solution. Instead of defining it as:
instance functorFoo :: Functor (Foo a) where
I need to define it as:
instance functorFoo :: Functor Foo where
But I'd still like to find out why does the first version not work. Like how does the Functor typeclass differ from e.g. Semigroup where the first version worked just fine.
answered Nov 22 '18 at 12:22
Rene SaarsooRene Saarsoo
9,05384567
add a comment |
I found a solution. Instead of defining it as:
instance functorFoo :: Functor (Foo a) where
I need to define it as:
instance functorFoo :: Functor Foo where
But I'd still like to find out why does the first version not work. Like how does the Functor typeclass differ from e.g. Semigroup where the first version worked just fine.
answered Nov 22 '18 at 12:22
Rene SaarsooRene Saarsoo
9,05384567
I found a solution. Instead of defining it as:
instance functorFoo :: Functor (Foo a) where
I need to define it as:
instance functorFoo :: Functor Foo where
But I'd still like to find out why does the first version not work. Like how does the Functor typeclass differ from e.g. Semigroup where the first version worked just fine.
answered Nov 22 '18 at 12:22
Rene SaarsooRene Saarsoo
9,05384567
answered Nov 22 '18 at 12:22
Rene SaarsooRene Saarsoo
9,05384567
answered Nov 22 '18 at 12:22
Rene SaarsooRene Saarsoo
9,05384567
answered Nov 22 '18 at 12:22
Rene SaarsooRene Saarsoo
9,05384567
9,05384567
add a comment |
add a comment |
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