module T1 = T2 does not result in (module T1) = (module T2) as types #5847
Comments
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Comment author: @alainfrisch 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. |
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Comment author: @zoggy Ok. Sorry, I missed it in the doc (I see it now). |
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Original bug ID: 5847
Reporter: @zoggy
Assigned to: @alainfrisch
Status: resolved (set by @alainfrisch on 2012-12-05T15:39:30Z)
Resolution: suspended
Priority: normal
Severity: minor
Version: 4.00.2+dev
Category: typing
Bug 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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