0005847: module T1 = T2 does not result in (module T1) = (module T2) as types - MantisBT

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ID

Project

Category

View Status

Date Submitted

Last Update

0005847

OCaml

OCaml typing

public

2012-12-05 14:25

2012-12-05 20:59

Reporter

guesdon

Assigned To

frisch

Priority

normal

Severity

minor

Reproducibility

always

Status

resolved

Resolution

suspended

Platform

OS

OS Version

Product Version

4.00.2+dev

Target Version

Fixed in Version

Summary

0005847: module T1 = T2 does not result in (module T1) = (module T2) as types

Description

The following code:

module type T1 = sig end

let foo t =
  let module T = (val t : T1) in
  ()
;;

module type T2 = T1

let bar (t : (module T2)) = foo t;;

gives the following error message:

File "mtype.ml", line 11, characters 32-33:
Error: This expression has type (module T2)
       but an expression was expected of type (module T1)

I expected T2 to be equal to T1 but it doesn't seem to the case. Am I missing something ?

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Relationships

Notes
(0008569) frisch (developer) 2012-12-05 16:39	The equality of packages types is based on the path equivalence of module types. In other words, we use nominal typing for first-class modules. This is the current expected behavior. Maybe it will be relaxed in the future, but there is no concrete plans to do so.

(0008572) guesdon (manager) 2012-12-05 20:59	Ok. Sorry, I missed it in the doc (I see it now).

Issue History
Date Modified	Username	Field	Change
2012-12-05 14:25	guesdon	New Issue
2012-12-05 15:04	gasche	Summary	module T1 = T2 does not result in T1 = T2 => module T1 = T2 does not result in (module T1) = (module T2) as types
2012-12-05 16:39	frisch	Note Added: 0008569
2012-12-05 16:39	frisch	Status	new => resolved
2012-12-05 16:39	frisch	Resolution	open => suspended
2012-12-05 16:39	frisch	Assigned To	=> frisch
2012-12-05 20:59	guesdon	Note Added: 0008572

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