How do I use a discriminated union branch in a type parameter?

Question

Suppose I have a type like this in F#:

type public Expression =
    | Identifier of string
    | BooleanConstant of bool
    | StringConstant of string
    | IntegerConstant of int
    | Vector of Expression list
    // etc...

Now I want to use this type to build a map:

definitions : Map<Identifier, Expression>

However, this gives the error:

The type 'identifier' is not defined

How can I use my type-case as a type paramter?

Answer 1

Identifier is a case constructor, not a type. It's actually a function with the type string -> Expression. The type of the case is string, so you can define definitions as

type definitions : Map<string, Expression>

Answer 2

There is another way in case you want the key to be a specific type (i.e.) not just another string. You can just create StringID type, and alternatively wrap that further into an Expression:

type StringId = Sid of string
type  Expression =
    | StringId of StringId
    | BooleanConstant of bool
    | StringConstant of string
    | IntegerConstant of int
    | Vector of Expression list

This will let you create a map in either of the following ways:

let x = Sid "x"
[StringId x ,BooleanConstant true] |> Map.ofList
//val it : Map<Expression,Expression> = map [(StringId (Sid "x"), BooleanConstant true)]

[x,BooleanConstant true] |> Map.ofList
//val it : Map<StringId,Expression> = map [(Sid "x", BooleanConstant true)]

That said, keeping the key as a simple string is certainly less complicated.