CODEWITHSUNDEEP
typesocamldeclaration
Ludovic Migneault
Ludovic Migneault

Reputation: 140

declaring a type that have no arguments

I have some types : 

type client =  {nom : nom_client; demande : demande_client}

type itineraire = {num : num_itineraire; 
                   capacite : capacite_itineraire; 
                   liste_clients : client list}

type plan = Vide | Ilist of itineraire list

I can declare clients and itineraires because I simply have to specify the arguments. However I don't know hoe to declare a variable to be of type plan.

type plan only contains a list of itineraire, but doing :

let myPlan = [(an itineraire here)];;

Returns an itineraire list instead of a plan.

So how do I declare a variable of Ilist of itineraire list?

Upvotes: 0

Views: 71

Answers (2)

gasche
gasche

Reputation: 31459

You should use the IList constructor, that takes an itineraire list as parameter as indicated in the type declaration, and returns a plan.

let my_plan = IList [foo; bar; baz]

Note that to get the list corresponding to a plan, you also need to pattern-match on this constructor.

let merge_plans p1 p2 = match p1, p2 with
  | p, Vide | Vide, p -> p
  | IList l1, IList l2 -> IList (l1 @ l2)

PS: do you really need to have this case distinction? Couldn't you defined plans simply as lists, and use the empty list instead of Vide?

Upvotes: 2

Jeffrey Scofield
Jeffrey Scofield

Reputation: 66818

It seems to me you just left out Ilist.

let myPlan = Ilist [];

Every value of type plan has either Vide or Ilist as its constructor. That's what the definition means.

Upvotes: 1

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