At 08:40 AM 4/15/98 +0200, Jean-Pierre Rosen wrote:
>Which raises an interesting question: how do you count:
>with A.B.C;
>
>as you know, it is equivalent to:
>with A, A.B, A.B.C;
>
>from the point of view of visibility rules, it is clearly equivalent to
three >withs. OTOH, it is likely that the programmer uses only the ultimate
child.

    Actually, you need to count them all, but only once.  I find myself in
Ada 95 appreciating the fact that if I have (for example) a child package
containing constructors, I have some with clauses that say:

    with Some_Abstraction;

    and others say:

    with Some_Abstraction.Creation;

    but often the spec of a package has the first, and the body the second.



                                        Robert I. Eachus

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