polymorphic variants and promotion
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Date:  20080219 (13:45) 
From:  Remi Vanicat <vanicat@d...> 
Subject:  Re: polymorphic variants and promotion 
"Johannes Kanig" <johannes.kanig@lri.fr> writes: > Hi, > > I ran into the following problem using polymorphic variants. Suppose I have > 2 functions f and g (the function definitions don't make much sense  they > are just there for typing purposes) : > > # let f (`A a) (`A i) = `A i ;; > val f : [< `A of 'a ] > [< `A of 'b ] > [> `A of 'b ] = <fun> > # let g (`A a) = (`B a) ;; > val g : [< `A of 'a ] > [> `B of 'a ] = <fun> > > For simplicity I assume that the type variables all instantiate to "int". > Now, I want to construct a function h that takes an accumulator a and a list > of `A's (for example [`A 1; `A 2; ...]) that produces either an object of > type `A of int or an object of type `B of int in the following way: > > # let h a = function >  [] > g a >  xs > List.fold_left f a xs;; > > But this doesn't type: > > This expression has type [ `A of 'a ] but is here used with type > [> `B of 'a ] > The first variant type does not allow tag(s) `B Well, you have to explicitly coerce : let h a = function  [] > g a  xs > (List.fold_left f a xs : [ `A of _ ] :> [> `A of _  `B of _ ]) ;; What I don't understand is why this : let h a = function  [] > g a  xs > (List.fold_left f a xs :> [> `A of _  `B of _ ]) ;; do not work ? I understood that we need to explicitly coerce, but the [ `A of _ ] is just the information the compiler will find by itself, why must we give it to it again ?  RÃ©mi Vanicat