Re: tagged type as generic parameter

[Georg Bauhaus <rm.tsoh+bauhaus@maps.futureapps.de>]

On Fri, 2008-01-04 at 14:08 +0100, Philippe Tarroux wrote:
> Dmitry A. Kazakov a écrit :
> >
> > The function A_Data returns Data. It is a primitive operation of Data and
> > thus covariant in the result. So when you derive anything from Data, you
> > have to override it in order to provide a correct implementation that would
> > return the derived type rather than the base.
> >   
> Ok. But does it mean that only the constructor is a primitive operation 
> of the data type, not the print procedure? In this case why Print 
> doesn't need to be redefined?

Data's Print operation works for objects of any type rooted at
Data. There is always a Data view of such an object.

OTOH, a function that is to return an object of a type
derived from Data will have to provide all components
of the derived type. That's not the LRM reason for the need
to override the function. But it should be a hint as to
why this need exists (the variance issue Dmitry has mentioned).

Imagine this

   Data
         function make (...) return Data;
    ^
    |  -- inheritance, IS A
  Special_Data


When you have a variable V of type Special_Data
and you call

  V := make (...);

then---without an overridden make---this Make would have to be
the Make defined with type Data. However, the Make defined
with Data cannot return objects of type Special_Data, which
is needed.

OTOH, When you have a variable V of type Special_Data
and you call

  print(V);

there is no problem because Print works with Data
objects and V "IS A" Data object (its type is derived
from Data.) So Print views V as being of type Data,
not Special_Data, in a sense.

HTH, Georg