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Why is this coercion necessary?
- Jacques Carette
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| Date: | -- (:) |
| From: | Jacques Carette <carette@m...> |
| Subject: | Why is this coercion necessary? |
Here is a much simplified version from a (much) larger problem I have
recently encountered:
type 'a a = [`A of 'a b]
and 'a b = [`B of 'a a]
and 'a c = [`C ]
type 'a d = [ 'a a | 'a b | 'a c]
type e = e d
# this code gives an error (details below)
let f1 (x:e) : e = match x with
| `A n -> n
| `B n -> n
| `C -> `C
# this works
let f2 (x:e) : e = match x with
| `A n -> (n :> e)
| `B n -> (n :> e)
| `C -> `C
f1 gives an error on the "| `B n -> n" line, pointing to the second 'n'
with
This expression has type e a but is used with type e b
These two variant types have no intersection
Indeed, they have no intersection, but they have a union! That is what
it seems the coercion in f2 'forces' the type-checker to realize, and
all works fine. But of course, such coercions end up polluting my code
all over the place (since the actual example is made of 9 types with 20
tags in total, and the 'recursive knot' requires 2 parameters to close
properly).
So, is this a bug? Is there a way to avoid these coercions?
Jacques