Re: [Caml-list] Strange Gadt error

[Jacques Garrigue <garrigue@math.nagoya-u.ac.jp>]

On 2015/10/12 23:20, Anders Peter Fugmann wrote:
> 
> Hi Jacques,
> 
> Thanks for detailed explanation. I think I understand now why the error occurs and more specifically how to fix it in a consistent way.
> 
> (However, changing 'add' to be
> 'add': int -> int -> int = fun n m-> n + m
> does not seem to help in my case)

Interesting.
This appears to be a rare case where it types with -principal, but not without it.
This is bug. I shall investigate it.

Jacques Garrigue

> 
> Thanks
> /Anders
> 
> On 09/10/15 01:17, Jacques Garrigue wrote:
>> On 2015/10/09 03:46, Anders Peter Fugmann wrote:
>>> 
>>> Hi,
>>> 
>>> I the following example (boiled down from a real use case):
>>> 
>>> type _ elem =
>>>  | Int: int elem
>>> 
>>> let rec incr: type a. a elem -> a -> int = function
>>>  | Int -> fun i -> add i 1
>>> and add n m = n + m
>>> 
>>> I get the error (Ocaml 4.02.3):
>>> File "example.ml", line 5, characters 24-25:
>>> Error: This expression has type int but an expression was expected of type
>>>         int
>>>       This instance of int is ambiguous:
>>>       it would escape the scope of its equation
>> 
>> Interesting error.
>> I see your confusion in seeing an error on ‘i’.
>> 
>> It is not completely wrong as you can indeed fix it by adding a local type
>> annotation changing the type of ‘i’ from ‘a’ to ‘int’.
>> 
>>> I can get rid of the error by annotating the type of i in line 5 like this:
>>> 
>>> | Int -> fun (i : int) -> add i 1
>>>                   ^^^
>> 
>> However, the real cause is not so much ‘i', whose type is indeed known (but as `a’, not `int’),
>> but rather the absence of type annotation on ‘add'.
>> Changing add in the following way fixes the problem:
>> 
>>   and add : int -> int -> int = fun n m -> n + m
>> 
>>> Or move add above incr like this:
>>> 
>>> let rec add n m = n + m
>>> and incr: type a. a elem -> a -> int = function
>>>  | Int -> fun i -> add i 1
>> 
>> This change of order only works by chance. If you use ocaml -principal, you still get
>> a type error here with this code.
>> 
>>> Is there an explanation to why I need to give the type of i in this case? As 'i' _must_ be an int (from the type annotation of incr), annotating the function seems ambiguous.
>> 
>> 
>> If you look carefully, you will see that the annotation says that ‘i’ has type ‘a’, not ‘int’.
>> In the local scope, those two types are equivalent, but once you leave if they are different.
>> Since we do not know yet the type of add, making a choice between the two seems arbitrary,
>> hence the error message.
>> 
>> The only conclusive source on how this works is my paper with Didier Rémy:
>> 	Ambivalent types for principal type inference with GADTs
>> 	http://pauillac.inria.fr/~remy/gadts/
>> 
>> In a nutshell, ambiguity occurs when a type obtained by unifying two equivalent
>> (but different) types is leaked out of their equivalence scope. What happens here is
>> a bit complicated. First the typer tries to give the type [a -> int -> int] to `add', avoiding
>> ambivalence. However, `a’ is not allowed to leak out of the definition of `incr’, so it
>> gets expanded into `int’. And this is that expansion which triggers the ambiguity
>> error. (An interesting remark is that, since add cannot have type [a -> int -> int] anyway,
>> there seems to be no ambiguity here. However, there is a scope between the
>> definition of `incr’ and the pattern-matching on `Int’ where such ambiguity might exists.)
>> By adding a local annotation on `i’, the problem is avoided, because then we are assuming
>> that `add’ has type [int -> int -> int] from the beginning (the ambivalence on `i’ does not leak).
>> Same thing with adding an annotation on `add’.
>> 
>> As specific remark on what happens when you change the order (without -principal):
>> Since add is typed first, and receives type [int -> int -> int], this type is handled as
>> though it was explicitly known when entering the gadt scope. This is done for
>> the type of all external identifiers, except for their non-generalized type variables.
>> As a result, you get the same behavior as adding a type annotation on add.
>> 
>> Thank you for this very demonstrative example.
>> 
>> Jacques Garrigue
>> 
>