Re: "access constant" discriminant

[mheaney@on2.com (Matthew Heaney)]

"Randy Brukardt" <randy@rrsoftware.com> wrote in message news:<v5d1sfo9ib2df9@corp.supernews.com>...
> Matthew Heaney wrote in message
> <1ec946d1.0302201642.66eb93e5@posting.google.com>...
> 
> >Here we see the reason why we passed an access parameter: it's because
> >we're returning an access value that designates one of our own
> >components.
> 
> 
> That reason being only that Ada doesn't support 'in out' parameters on
> functions. If it did,
> 
> fuction Persistence_View
>      (Object : in out T) return Persistence_Class_Access is
>    begin
>       return Object.Persistence_View'Access;
>    end;
> 
> would work fine, and would be a lot easier to use as well.

No -- this would be the wrong way to implement the multiple views
idiom.  You really do want to make explicit the fact that the object
is being aliased.

For example, in C++ I could do this:

  X x;
  Y y(x);

and implement Y this way:

class Y
{
   X& x;
public:
   Y(X& xx) : x(xx) {}
   //...
private:
   Y(const Y&);
   Y& operator=(const Y&);
};

but that would be entirely misleading, even if it's "easier" to
declare a Y object.  It's better to implement Y's ctor using an
explicit pointer:

class Y
{
   X& x;
public:
   Y(X* px) : x(*px) {}
private:
   Y(const Y&);
   Y& operator=(const Y&);
};

and then say:

  X x;
  Y y(&x);

Which makes it identical to how you'd say it in Ada95.  So this is not
really an Ada issue.

 
> But that's not going to happen, so we're stuck with terrible workarounds
> like 'access' parameters.

Access parameters have other uses besides getting around the lack of
inout mode for function parameters.  For example, operations that
accept a user-defined type as an access parameter are primitive for
the type, and hence are inherited in a derivation.  And an access
parameter means you don't have to say Op (O.all) everywhere.

However, we are in agreement that not having inout mode for functions
is a ridiculous restriction, since functions in Ada95 do modify their
arguments (the language just doesn't let you *say* that you do).