Abstraction in SML 
Author Message
 Abstraction in SML

Hi,

I've got some trouble to understand abstraction in SML/NJ 0.93.
To learn , I'm using " Introduction to SML" by Ropert Harper.
He said that if we define:

signature SIG =
sig
    type t
    val x: t -> t
end;

and

abstraction V:SIG =
struct
    type t = int
    val x = fn x => x
end;

First, he said that the given result of the previous definition is:

Quote:
> abstraction V:SIG
or I obtain
> structure V:SIG

and that we should be enable to do:
V.x(3);
but I've got the error:
std_in:18.1-18.6 Error: operator and operand don't agree (tycon
mismatch)
  operator domain: V.t
  operand:         int
  in expression:
    x (3)

What's wrong with it ?
Moreover, can somebody tell me more about abstraction ?

Thanks,
        Stephane Baudet
--




Tue, 03 Nov 1998 03:00:00 GMT  
 Abstraction in SML

Those notes are somewhat dated and should be revised.

The effect of an abstraction declaration (which is not official SML,but
is supported by SML/NJ) is to restrict type sharing information to
exactly that which is specified in the signature.  So if you make an
abstraction binding and the signature merely says "type t", then this
type is distinct from all other types.

As you can readily see, abstraction bindings are not always what you
would like.  eg, if we make an abstraction A of signature POSET, where

signature POSET = sig

        type t
        val le : t * t -> bool

end

then A is useless because A.le can never be applied!  The solution to
this problem is to provide a finer degree of control over visibility of
type information by allowing type definitions to occur in signatures.

A planned revision of SML (to be completed in about a month) will
provide a more flexible variant of abstraction bindings and admit type
definitions in signatures, leading to a significantly more flexible
module system.

Bob Harper



Fri, 06 Nov 1998 03:00:00 GMT  
 
 [ 2 post ] 

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