Type indexed types?
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Date:  20050125 (17:01) 
From:  skaller <skaller@u...> 
Subject:  Re: [Camllist] Type indexed types? 
On Tue, 20050125 at 23:28, Jim Farrand wrote: > Hi, > > This is what I need to do: > > Given a type, u, and a collection of types, ts, find a type t such that: > * t is a member of ts > * t can be unified with u > * There is no type v in ts, which unifies with u and is more general > than t. > > (In other words, find the most specific type in ts that is still equal > to or a generalisation of u.) This is the overload resolution problem? Shouldn't that last condition be reversed? You're looking for the most specialised, not most general type that matches u. > For example: > > ts = list int, int, list 'a, 'a > > For u = list int, the result would be list int. For u = list string, > the result would be list 'a, and for u = string the result would be 'a. > > Note that I already have a function which can compute if type a is a > specialisation of type b. > > I think that in theory, I could do a brute force approach: > * Grab all types in ts that are generalisations of u. > * Order the results according to how specific they are. > * Arbitrarily choose one of the most specific. > > But there must be a smarter way! Not really. Here is what Felix does: first, find all the matches. Now we need the most specialised. My notes say: (* start with an empty list, and fold one result at a time into it, as follows: if one element of the list is greater (more general) than the candidate, then add the candidate to the list and remove all element greater than the candidate, otherwise, if one element of the list is less then the candidate, keep the list and discard the candidate. The list starts off empty, so that all elements in it are vacuously incomparable. It follows either the candidate is not less than all the list, or less than all the list: that is, there cannot be two element a,b such that a < c < b, since by transitivity a < c would follow, contradicting the assumption the list contains no ordered pairs. If in case 1, all the greater element are removed and c added, all the elements must be less or not comparable to c, thus the list remains without comparable pairs, otherwise in case 2, the list is retained and c discarded and so trivially remains unordered. *) So basically, you keep a list of incomparable types, and fold one of the candidates into it at a time, retaining the invariant. If the candidate compares with any element in the list, you either chuck out the candidate because it is too general (the list remains the same), or you chuck out all the elements in the list that it is comparable with, and then add it to the list (so the list ends up with all incomparable elements again). Here's my actual algorithm, cleaned up a bit. As far as I know this is the best possible algorithm. let candidates = fold_left (fun oc r > match r with Unique (j,c,_,_) > let rec aux lhs rhs = match rhs with  [] > r::lhs (* return all nongreater elements plus candidate *)  (Unique(i,typ,mgu,ts) as x)::t > begin match compare_sigs syms.dfns typ c with  `Less > lhs @ rhs (* keep whole list, discard c *)  `Equal > ERROR "[resolve_overload] Ambiguous call"  `Greater > aux lhs t (* discard greater element *)  `Incomparable > aux (x::lhs) t (* keep element *) end  Fail::_ > assert false in aux [] oc  Fail > assert false ) [] candidates in match candidates with  [Unique (i,t,mgu,ts)] > Some (i,t,mgu,ts)  [] > None  _ > ERROR "More than one match"  John Skaller, mailto:skaller@users.sf.net voice: 061296600850, snail: PO BOX 401 Glebe NSW 2037 Australia Checkout the Felix programming language http://felix.sf.net