What's "node" doing in this snippet of SML?

Question

I'm working on a project that has a lot code that looks like this:

structure MyStruct = struct
datatype node
  = A of Foo.t
  | B

type t = node Wrap.t

fun layout myNode =
  case node myNode of
    A foo => Foo.bar foo 
  | B => "void"
end

Wrap is defined elsewhere, but the main point is that it has one parametrized type, t.

What is node doing in the line case node myNode of? It's not a type constructor, since t isn't a type, and changing it to case (node myNode) throws an error message for that reason. Additionally, if I write functions that look like this:

fun layout node myNode = ...

I get a type error because it doesn't have the type MyStruct.t -> Layout.t but rather 'a -> MyStruct.node -> Layout.t. If I do the same thing inside of a case statement, however, it's fine.

What's going on here?

Answer 1

There must be some function named node defined in the program, whose type is u -> node, for some u. That function is applied here. From that, the layout function then is simply inferred to have type u -> string.