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| Date: | -- (:) |
| From: | rossberg@m... |
| Subject: | Re: [Caml-list] Implicitely abstracted type |
Alain Frisch wrote: > > When applying a functor of type > functor(X:S1) -> S2 to a module of type T, the module type for the > result can be obtained in two different ways: > > (1) T is a path: the module type is obtained by substituting X with T in S2. > > (2) T is not a path: the module type is obtained by computing the > smallest supertype of S2 that doesn't contain X anymore (under the extra > assumption that X has type T). I believe this doesn't type-check. :) You probably meant to say: "When applying a functor of type functor(X:S1) -> S2 to a module M of type T, the module type for the result can be obtained in two different ways: (1) M is a path: the module type is obtained by substituting X with M in S2. (2) M is not a path: the module type is obtained by computing the smallest supertype of S2 that doesn't contain X anymore (under the extra assumption that X has type T)." /Andreas