Re: Type invariants and private extensions?

[Jere <jhb.chat@gmail.com>]

On Wednesday, September 13, 2017 at 12:21:12 AM UTC-4, Jere wrote:
> On Tuesday, September 12, 2017 at 6:59:17 PM UTC-4, Victor Porton wrote:
> > Shark8 wrote:
> > 
> > > On Tuesday, September 12, 2017 at 2:09:08 PM UTC-6, Victor Porton wrote:
> > >> Jeffrey R. Carter wrote:
> > >> 
> > >> > 
> > >> > "I want to" is not the same as "there is no way to solve the problem in
> > >> > current Ada". Ada has a feature that provides exactly what you need.
> > >> > It's called a variant record.
> > >> 
> > >> It is a tagged type. AFAIK, a type cannot be both a variant record and
> > >> tagged.
> > > 
> > > You certainly can:
> > > 
> > >     Type Type_Enumeration is ( TInteger, TReal, TBoolean, Nothing );
> > >     
> > >     Type Example( Item_Type : Type_Enumeration ) is tagged record
> > > case Item_Type is
> > > when TInteger => I : Integer;
> > > when TReal    => R : Float range Float'Range;
> > > when TBoolean => B : Boolean;
> > > when Nothing  => Null;
> > > end case;
> > >     end record;
> > >     
> > >> Moreover, the object in consideration is a wrapper over a certain C API.
> > >> It surely cannot be described in variant record terms.
> > > 
> > > Sure it can; there's a reason that there's a separation between
> > > specification and body, part of which is so you can hide something like an
> > > interface to C and present something sensible to your program/clients.
> > 
> > Thanks.
> > 
> > But this seems not to solve my problem:
> > 
> > The base type for Example cannot be defined as a discriminated type for 
> > certain Type_Enumeration, because it is possible that when creating the 
> > object it may be yet unknown what the value of the discriminant should have 
> > (my main problem is to invent somethings if this value is known at object 
> > creation, but we must support the unknown case too). AFAIK, it is not 
> > possible to change the discriminant later.
> > 
> 
> I may be misunderstanding what you are looking for, but you can 
> make discriminant types unconstrained to let you choose the type 
> at run time:
> 
> Type Type_Enumeration is ( TInteger, TReal, TBoolean, Nothing );
>    
>     Type Example( Item_Type : Type_Enumeration := Nothing) is tagged record
>         case Item_Type is
>             when TInteger => I : Integer;
>             when TReal    => R : Float range Float'Range;
>             when TBoolean => B : Boolean;
>             when Nothing  => Null;
>         end case;
>     end record; 
> 
> Then all you have to do is declare an object with no 
> specified discriminant:
> 
>  Some_Object : Example;  --notice no discriminant
> 
> In Ada, when you do this, you are able to assign new versions 
> of the type (with different discriminants:
> 
> Some_Integer : Example := (Item_Type => TInteger, I => 100);
> Some_Float   : Example := (Item_Type => TFloat,   F => 20.2);
> 
> Some_Object := Some_Integer;
> Some_Object := Some_Float;
> 
> That's just a play example, you can handle building them however 
> you want at run time.

Sorry, TFloat should be TReal and F should be R

Also, I didn't realize this, but you can't do the default value with a tagged type unless it is limited so it would need to be untagged:

   Type Example( Item_Type : Type_Enumeration := Nothing) is record
        case Item_Type is
            when TInteger => I : Integer;
            when TReal    => R : Float range Float'Range;
            when TBoolean => B : Boolean;
            when Nothing  => Null;
        end case;
    end record; 

If you are selecting the type via enum, does it still need to be tagged at this point?