Data types basics

[Patty Fong]

Another prompt reply! thanks Hal :) I think I understand this correctly now

For my previous problem which was:
<prolog> ::= (<assertion> ".")*
<assertion> :: = <structure> | <rule>
<rule> ::= <structure> ":-" <structure>("," <structure>)*
<structure> ::= <name> [“(“ <term> (“,” <term>)* “)”]
<term> ::= <number> | <variable> | <structure>
<variable> ::= <name>

Does this appear to be correct:
type Prolog = [Assertion]
data Assertion = StrucAssert Structure | RuleAssert Rule
data Rule = Rule Structure Structure [Structure]
data Structure = Structure Name Term [Term]
data Term = NumTerm Number | VarTerm Variable | StrucTerm Structure
type Variable = Name
type Name = String
type Number = Int

>From: Hal Daume III To: Patty Fong CC: Haskell Cafe , Hal Daume Subject: 
>Re: Data types basics Date: Tue, 4 Nov 2003 22:10:03 -0800 (PST)
>
>Hi again,
>
>On Wed, 5 Nov 2003, Patty Fong wrote:
>
> > Hi to anyone reading this. i'm still strugling a bit with data type > 
>declarations. > > The was i understand it is that if i delcare a new data 
>type: > > data myType = myType a | b | c
>
>This isn't entirely correct. The names of types have to begin with capital 
>letters, as do constructors (more later). So you would need this to be:
>
>data MyType = MyType A | B | C
>
>where A is an existing type.
>
>This type now has three constructors:
>
>MyType :: A -> MyType B :: MyType C :: MyType
>
>It's perhaps a bit easier to understand when the names are different.
>
>When we say:
>
>data Foo = Bar Int | Baz String
>
>this means that a "Foo" is of one of two forms (the "|" can be read as 
>disjunction). A value of type Foo is either of the form "Bar x" for some x 
>which is an Int or "Baz y" for some y which is a String.
>
>"Bar" and "Baz" are called "constructors" because they take arguments and 
>"construct" a Foo. So, in this case,
>
>Bar :: Int -> Foo Baz :: String -> Foo
>
>are the two constructors.
>
>Of course, you have have any number of constructors and each can have any 
>number of arguments.
>
>data Foo = Bar | Baz Int | Bazaa String Bool
>
>Now there are three constructors:
>
>Bar :: Foo Baz :: Int -> Foo Bazaa :: String -> Bool -> Foo
>
>they can be recursive:
>
>data Foo = Bar | Baz Int Foo
>
>Bar :: Foo Baz :: Int -> Foo -> Foo
>
>and can have "type variables", for instance:
>
>data Foo a = Bar | Baz a
>
>here, something of type "Foo a" is either of the form Bar or of the form 
>Baz x for some x which is of type a. This has constructors:
>
>Bar :: Foo a Baz :: a -> Foo a
>
>I hope this sheds some light on the issue...
>
>
>--
>Hal Daume III | hdaume at isi.edu "Arrest this man, he talks in maths." | 
>www.isi.edu/~hdaume
>

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