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How to create ADT in Haskell?

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16

2

In Scala I can describe such ADT:

sealed trait Foo
case class A(a: Int) extends Foo
case class B(b: String) extends Foo
case class C(a: A, b: B) extends Foo

How can I do the same in Haskell?

data Foo = A Int | B String | C A B

It doesn't work, because A and B are not types. Should I use GHC extensions to do it?

scala haskell algebraic-data-types

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edited yesterday

mfirry

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AndrewAndrew

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16

2

In Scala I can describe such ADT:

sealed trait Foo
case class A(a: Int) extends Foo
case class B(b: String) extends Foo
case class C(a: A, b: B) extends Foo

How can I do the same in Haskell?

data Foo = A Int | B String | C A B

It doesn't work, because A and B are not types. Should I use GHC extensions to do it?

scala haskell algebraic-data-types

share|improve this question

edited yesterday

mfirry

2,75811729

asked yesterday

AndrewAndrew

834

New contributor

Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

In Scala I can describe such ADT:

sealed trait Foo
case class A(a: Int) extends Foo
case class B(b: String) extends Foo
case class C(a: A, b: B) extends Foo

How can I do the same in Haskell?

data Foo = A Int | B String | C A B

It doesn't work, because A and B are not types. Should I use GHC extensions to do it?

scala haskell algebraic-data-types

share|improve this question

edited yesterday

mfirry

2,75811729

asked yesterday

AndrewAndrew

834

New contributor

Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

sealed trait Foo
case class A(a: Int) extends Foo
case class B(b: String) extends Foo
case class C(a: A, b: B) extends Foo

data Foo = A Int | B String | C A B

share|improve this question

edited yesterday

mfirry

2,75811729

asked yesterday

AndrewAndrew

834

New contributor

Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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edited yesterday

mfirry

2,75811729

asked yesterday

AndrewAndrew

834

New contributor

Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited yesterday

mfirry

2,75811729

edited yesterday

mfirry

2,75811729

asked yesterday

AndrewAndrew

834

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Check out our Code of Conduct.

asked yesterday

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26

In Scala, your ADT makes A,B,C to be subtypes of Foo. In Haskell we do not have subtypes, so A,B,C are instead constructors of type Foo.

A few possible workarounds:

1. 
   

   Repeat the fields. This is the most basic option.

   
   
   

   data Foo = A Int | B String | C Int String
   

   
   


2. 
   

   Define additional types, so that we can reuse them more than once.

   
   
   

   data AT = AT Int -- can have many arguments
   data BT = BT String -- can have many arguments
   data Foo = A AT | B BT | C AT BT
   

   
   


3. 
   

   Exploit a GADT

   
   
   

   data FooTag = AT | BT | CT
   
   data Foo (tag :: FooTag) where
    A :: Int -> Foo 'AT
    B :: String -> Foo 'BT
    C :: Foo 'AT -> Foo 'BT -> Foo 'CT
   

   
   
   

   Here, in the last line we are able to refer to "the values constructed using A" using the type Foo 'AT, since tag AT is only used by constructor A.
   Note that this approach adds a tag parameter to Foo, so it slightly changes the interface: we can no longer write bar :: Foo -> ..., but we have to write bar :: Foo t -> ... (or to use existential types).

   
   

share|improve this answer

edited yesterday

answered yesterday

chichi

76.8k287145

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  The last example requires several extensions, though, right? (DataKinds, GADTs, and KindSignatures, I think.)
  
  – chepner
  yesterday
  

  
  
  
  

add a comment | 

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26

In Scala, your ADT makes A,B,C to be subtypes of Foo. In Haskell we do not have subtypes, so A,B,C are instead constructors of type Foo.

A few possible workarounds:

1. 
   

   Repeat the fields. This is the most basic option.

   
   
   

   data Foo = A Int | B String | C Int String
   

   
   


2. 
   

   Define additional types, so that we can reuse them more than once.

   
   
   

   data AT = AT Int -- can have many arguments
   data BT = BT String -- can have many arguments
   data Foo = A AT | B BT | C AT BT
   

   
   


3. 
   

   Exploit a GADT

   
   
   

   data FooTag = AT | BT | CT
   
   data Foo (tag :: FooTag) where
    A :: Int -> Foo 'AT
    B :: String -> Foo 'BT
    C :: Foo 'AT -> Foo 'BT -> Foo 'CT
   

   
   
   

   Here, in the last line we are able to refer to "the values constructed using A" using the type Foo 'AT, since tag AT is only used by constructor A.
   Note that this approach adds a tag parameter to Foo, so it slightly changes the interface: we can no longer write bar :: Foo -> ..., but we have to write bar :: Foo t -> ... (or to use existential types).

   
   

share|improve this answer

edited yesterday

answered yesterday

chichi

76.8k287145

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  The last example requires several extensions, though, right? (DataKinds, GADTs, and KindSignatures, I think.)
  
  – chepner
  yesterday
  

  
  
  
  

add a comment | 

26

In Scala, your ADT makes A,B,C to be subtypes of Foo. In Haskell we do not have subtypes, so A,B,C are instead constructors of type Foo.

A few possible workarounds:

1. 
   

   Repeat the fields. This is the most basic option.

   
   
   

   data Foo = A Int | B String | C Int String
   

   
   


2. 
   

   Define additional types, so that we can reuse them more than once.

   
   
   

   data AT = AT Int -- can have many arguments
   data BT = BT String -- can have many arguments
   data Foo = A AT | B BT | C AT BT
   

   
   


3. 
   

   Exploit a GADT

   
   
   

   data FooTag = AT | BT | CT
   
   data Foo (tag :: FooTag) where
    A :: Int -> Foo 'AT
    B :: String -> Foo 'BT
    C :: Foo 'AT -> Foo 'BT -> Foo 'CT
   

   
   
   

   Here, in the last line we are able to refer to "the values constructed using A" using the type Foo 'AT, since tag AT is only used by constructor A.
   Note that this approach adds a tag parameter to Foo, so it slightly changes the interface: we can no longer write bar :: Foo -> ..., but we have to write bar :: Foo t -> ... (or to use existential types).

   
   

share|improve this answer

edited yesterday

answered yesterday

chichi

76.8k287145

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  The last example requires several extensions, though, right? (DataKinds, GADTs, and KindSignatures, I think.)
  
  – chepner
  yesterday
  

  
  
  
  

add a comment | 

In Scala, your ADT makes A,B,C to be subtypes of Foo. In Haskell we do not have subtypes, so A,B,C are instead constructors of type Foo.

A few possible workarounds:

1. 
   

   Repeat the fields. This is the most basic option.

   
   
   

   data Foo = A Int | B String | C Int String
   

   
   


2. 
   

   Define additional types, so that we can reuse them more than once.

   
   
   

   data AT = AT Int -- can have many arguments
   data BT = BT String -- can have many arguments
   data Foo = A AT | B BT | C AT BT
   

   
   


3. 
   

   Exploit a GADT

   
   
   

   data FooTag = AT | BT | CT
   
   data Foo (tag :: FooTag) where
    A :: Int -> Foo 'AT
    B :: String -> Foo 'BT
    C :: Foo 'AT -> Foo 'BT -> Foo 'CT
   

   
   
   

   Here, in the last line we are able to refer to "the values constructed using A" using the type Foo 'AT, since tag AT is only used by constructor A.
   Note that this approach adds a tag parameter to Foo, so it slightly changes the interface: we can no longer write bar :: Foo -> ..., but we have to write bar :: Foo t -> ... (or to use existential types).

   
   

share|improve this answer

edited yesterday

answered yesterday

chichi

76.8k287145

share|improve this answer

edited yesterday

answered yesterday

chichi

76.8k287145

answered yesterday

chichi

76.8k287145

answered yesterday

chichi

76.8k287145