Re: static virtual is a keyhole?

> >In C++ we have cv qualifiers, const, volatile, both, neither. The current
> >language allows all the combinations.
> >
> >Suppose everything were as is, just const-volatile were not allowed.
Would
> >you call that a (gratuitous) keyhole applying your definition?
>
> Probably not.  To date, I have resisted the urge to classify missing
features as keyholes.  The kinds of keyholes I'm interested in are
gratuitous restrictions on features that are offered.  For example, if some
programming language said you can't have an array with more than 100
elements, I'd call that a keyhole.  But if the language failed to offer
arrays, the lack of arrays would not be a keyhole.

Scott, I tried to figure out the logic here, but failed.   So the most
likely solution I found is you misimterpreted my example.  Let me restate
it:

Current language:

0)   int;        // nonvolatile, nonconst
1)  const int;  // nonvolatile, const
2)  volatile int;  // volatile, nonconst
3)  const volatile int;  // volatile, const

Supposed language variants:

A)  remove cases 1, 3:   means no const.  classification: "missing feature"
B)  remove cases 2, 3:   means no volatile.  classification: "missing
feature"
C)  remove cases 1, 2, 3:    classification: "missing feature"   (I suppose
you interpreted my words this way)
D)  remove only case 3:    my classification: "keyhole"   (my original
meaning)

As here the language has to implement both the features, and I see  no sane
reason to not allow the combination of the otherwise disjunct features.

That well converts to your array example.  Suppose our target is a 2-bit
processor.  And support was implemented, but with range limited to [0..2],
as opposed to the limitless coverage of the full addressable range of
[0..3].

If you thought of case D can you see why we come to different conclusions
following seemingly the same rationale?

Paul