Re: type sharing
 trc@i...
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Date:  19980316 (15:04) 
From:  trc@i... 
Subject:  Re: type sharing 
I find a getaround for the problem I posed eariler. I wonder if there are other ways to get around it. best wishes, TyngRuey  Now the last line of the following ocaml code will typecheck. But if you remove the "with module Quad = B.Quad" in the specification of functor Dtree, it won't. That is exactly what I want. ===== start Ocaml code ===== module type QUAD = sig type ('a, 'b) t = AA of 'a * 'a  AB of 'a * 'b  BA of 'b * 'a  BB of 'b * 'b end module Quad: QUAD = struct type ('a, 'b) t = AA of 'a * 'a  AB of 'a * 'b  BA of 'b * 'a  BB of 'b * 'b end module type BTREE = sig module Quad: QUAD type ('a, 'b) quad = ('a, 'b) Quad.t end module Btree(Q: QUAD): BTREE = struct module Quad = Q type ('a, 'b) quad = ('a, 'b) Quad.t end module Dtree(B: BTREE): BTREE with module Quad = B.Quad = struct module Quad = B.Quad type ('a, 'b) quad = ('a, 'b) B.quad end module B = Btree(Quad) module D = Dtree(B) let b = B.Quad.AA (1, 1) let d = D.Quad.AA (1, 1) let u = b = d ==== end Ocaml code === > From trc@ccs1 Sun Mar 15 14:33:05 1998 > To: camllist@inria.fr > Cc: trc@ccs1 > > Hello, > > I am trying to translate some SML modules with type sharing constraints > into corresponding Ocaml modules but without success. > I wonder if there is a good fix. Thanks in advance. > > best wishes, > > TyngRuey Chuang > >  > > Here is the detailed description. > > In SML, one can do the following: > > ===== start SML code ===== > signature BTREE > = > sig > datatype ('a, 'b) quad = AA of 'a * 'a >  AB of 'a * 'b >  BA of 'b * 'a >  BB of 'b * 'b > end > > > structure Btree:> BTREE > = > struct > datatype ('a, 'b) quad = AA of 'a * 'a >  AB of 'a * 'b >  BA of 'b * 'a >  BB of 'b * 'b > end > > > functor Dtree(B: BTREE):> BTREE where type ('a, 'b) quad = ('a, 'b) B.quad > = > struct > open B > end > > structure B = Btree > structure D = Dtree(B) > > val u = B.AA (1, 1) = D.AA (1, 1) > ===== end SML code ===== > > The last line of code will typecheck. If the "where ..." clause > in the specification of functor Dtree is deleted, then the same line > will not typecheck. I think this is quite nice as it allows one to > control the exposure of type sharing information among the modules. > I intend to keep this mechanism in my Ocaml translation. > > The direct translation of the above SML code to Ocaml is: > > ==== start Ocaml code ===== > module type BTREE > = > sig > type ('a, 'b) quad = AA of 'a * 'a >  AB of 'a * 'b >  BA of 'b * 'a >  BB of 'b * 'b > end > > > module Btree: BTREE > = > struct > type ('a, 'b) quad = AA of 'a * 'a >  AB of 'a * 'b >  BA of 'b * 'a >  BB of 'b * 'b > end > > > module Dtree(B: BTREE): BTREE with type ('a, 'b) quad = ('a, 'b) B.quad > = > struct > open B > end > > module B = Btree > module D = DistBtree(B) > > let b = B.AA (1, 1) > let d = D.AA (1, 1) > let u = b = d > ==== end Ocaml code ==== > > This will not work because "open" has a different semantics in Ocaml > than in SML. The best I can do so far is the following (where > type quad in module type BTREE becomes abstract): > > ==== start Ocaml code ==== > module type BTREE > = > sig > type ('a, 'b) quad > > val aa: 'a * 'a > ('a, 'b) quad > end > > > module Btree: BTREE > = > struct > type ('a, 'b) quad = AA of 'a * 'a >  AB of 'a * 'b >  BA of 'b * 'a >  BB of 'b * 'b > let aa (u, v) = AA (u, v) > end > > > module Dtree(B: BTREE): BTREE with type ('a, 'b) quad = ('a, 'b) B.quad > = > struct > type ('a, 'b) quad = ('a, 'b) B.quad > > let aa (u, v) = B.aa (u, v) > end > > > module B = Btree > module D = Dtree(B) > > let b = B.aa (1, 1) > let d = D.aa (1, 1) > let v = b = d > ===== end Ocaml code ==== > > But now I lose the nice constructors AA, etc. (I can no longer > patternmatch them.) > > Is there an easy fix to this problem? Many thanks! > > > > >