AST类型可以在ocaml中递归吗？

Question

I have attempted to translate my grammar into an AST.

Can an AST type be recursive? For instance, I have a production eprime -> PLUS t eprime | MINUS t eprime | epsilon eprime -> PLUS t eprime | MINUS t eprime | epsilon eprime -> PLUS t eprime | MINUS t eprime | epsilon . Is it correct to translate that to:

type eprime = 
| Add of t eprime 
| Minus of t eprime 
| Eempty

Answer 1

The short answer is yes. This is more or less exactly how you define a tree-shaped data structure.

A syntactically correct definition looks more like this:

type eprime = 
| Add of t * eprime 
| Minus of t * eprime 
| Empty

If you assume t is int (for simplicity), you can create a value of this type like this:

# Add (3, Add (4, Empty));;
- : eprime = Add (3, Add (4, Empty))

Answer 2

Yes, an AST type can be recursive and often is. However the correct syntax would be Add of t * eprime . Without the * the t would be seen as a type argument to eprime , which doesn't take any.

PS: You don't have to (and probably shouldn't) model your AST after your grammar as closely as you do. It is perfectly okay to have "left recursion" in the AST, even if you've removed it from your grammar. Similarly you don't have to encode operator precedence in your AST types the same way you do in the grammar, so for example having Add and Mult in the same type is no problem. With that in mind the usual definition of an AST for expressions looks more like this:

type exp =
  | Add of exp * exp
  | Sub of exp * exp
  | Mult of exp * exp
  | Div of exp * exp
  | FunctionCall of ident * exp list
  | Var of ident
  | Const of value