Re: Reemergence of predefined equality operator...

[Robert A Duff <bobduff@shell01.TheWorld.com>]

Dale Stanbrough <dstanbro@bigpond.net.au> writes:

> I was demonstrating what i thought would be the reemergence of
> the predefined equality operator by doing...
> 
>    type text is... (record with length & chars, like bounded string)
>    function "=" (a, b : text) return boolean;
> 
> 
> and
> 
> 
>    generic
>       type element is private;
>       with function "=" (a, b : element) return boolean is <>;
>    package...
> 
> and after instantiating the generic with type text it kept calling
> the "=" defined for type text.
> 
> I thought Ada (through a bad design decision) said that it should call
> the predefined "=" opertor in this instance. Has this been changed,
> or do have I misunderstood the problem?

You've misunderstood the problem.  ;-)

Above, you passed in a formal "=" explicitly.  The "is <>" means that
the default is to pass in whatever "=" happens to be visible at the
place of the instantiaation (and has the right profile types).  That is
probably the user-defined "=" that you wanted.  In that case, the
"problem" does not occur.

The "problem" is when you leave out the explicit ``with function
"="...''.  Then type "element" has a predefined "=" (because it's
private, not limited private).  In that case, the predefined "=" of the
actual type "reemerges".  In other words, if you instantiate with
Element => T, and T has a primitive user-defined "=", the instance will
call the predefined "=" rather than the user-defined one.

If you wanted to drive a stake through the heart of the predefined "="
by declaring a user-defined "=", so that every time objects of type T
are compared for equality it will call the user-defined "=", then you
lose -- if you pass T to a generic or wrap T in a record, the predefined
"=" reemerges (rises from the dead).

For "=", this seems "obviously" wrong.  But as I pointed out elsewhere,
the situation for other operators (eg, "mod"), is not so obvious.
Even if the language were changed, or even if Text_IO.Integer_IO took
"mod" as an "is <>" parameter as above, it's not clear (to me) what the
right answer is.

- Bob