Shadowing types in OCaml

updated: 2025-11-17

One weird thing in OCaml is that like variables, you can shadow types.

Shadowing types is NOT the same as overriding a type.

Here is an example:

let x = 3
type int = Type
let y = Type
let z = 10
let w : int = 2

Here, the original int type still technically exists, but you just can't access it through the type int anymore. z can still be defined, because OCaml can still infer it's got the original int type, but we can't define w like that, because the type annotation for int now means something else.

It can get funky in utop

Utop can show some weird things when you shadow types. I will overshadow ints again as an example. Check out the following experiment in utop. The curious thing is that here, since the original int type got overshadowed twice, typing in an integer will say it's got type int/3, because it's matches the type of the third to last int definition.

──────────┬─────────────────────────────────────────────────────────────┬───────────
          │ Welcome to utop version 2.16.0 (using OCaml version 5.3.0)! │
          └─────────────────────────────────────────────────────────────┘

Type #utop_help for help about using utop.

─( 00:11:41 )─< command 0 >──────────────────────────────────────────{ counter: 0 }─
utop # type int = A;;
type int = A
─( 00:11:41 )─< command 1 >──────────────────────────────────────────{ counter: 0 }─
utop # let a = A;;
val a : int = A
─( 00:11:56 )─< command 2 >──────────────────────────────────────────{ counter: 0 }─
utop # type int = B;;
type int = B
─( 00:11:59 )─< command 3 >──────────────────────────────────────────{ counter: 0 }─
utop # let b = B;;
val b : int = B
─( 00:12:03 )─< command 4 >──────────────────────────────────────────{ counter: 0 }─
utop # a;;
- : int/2 = A
─( 00:12:06 )─< command 5 >──────────────────────────────────────────{ counter: 0 }─
utop # 10;;
- : int/3 = 10
─( 00:12:07 )─< command 6 >──────────────────────────────────────────{ counter: 0 }─