Re: Default equality for generic formal types

["R. Tim Coslet" <coslett@kaisere.com>]

See AARM 12.3(29.b)

For upward compatibility with Ada83 (which had this bug/feature) for
untagged types the the default function reverts to the predefined one. I
can't seem to locate the reference right now, but I believe that even if you
explicitly passed the instantion the "=" function, it still reverts to the
predefined one. However if your "=" function had a different name (like
"equal" you could pass that explicitly (a rename might work to accomplish
this workaround, but if not a wrapper function will).

This is not a problem with tagged types.

Matthew Heaney wrote:

> Suppose I have a generic subprogram like this:
>
> generic
>   type Item_Type is private;
>   with function Item_Eq (L, R : Item_Type)
>     return Boolean is "=";
> procedure Generic_Op (O : T);
>
> Suppose further that the generic actual type is nontagged, and has
> overridden its predefined equality.
>
> To which equality does Item_Eq refer?  The overridden version, or the
> predefined version?  What does that default value for Item_Eq, "=", mean
> at the time of compilation of the generic?
>
> If the client does not explicitly pass in an item equality op during the
> instantiation (likely, because the type is nonlimited), then inside
> Generic_Op, what does Item_Eq refer to?  Predefined equality, or the
> overridding version?
>
> Thanks in advance,
> Matt