On Monday, 21 December 2015 at 20:29:14 UTC, Steven Schveighoffer
wrote:
1) Is this recursion expected?
Yes. assert calls the virtual invariant function, which in the
case of super is equivalent to this. So you are essentially
calling assert(this).
2) The example is a dustmite'd version of this:
It seems like something you shouldn't do. AFAIK, invariant
should not call any public functions on your existing class.
Thanks for the reply! Yeah, this helps for a clearer picture of
what's happening.
In particular, even though all invariants of a class hierarchy
are tested instead of only the most-subclassed-one, triggering
the invariant check remains virtual. I didn't know that.
Even if Base.f() is const, it's not allowed inside
Derived.invariant(). This is again understandable: By OOP
principles, Derived shouldn't impose further restrictions on Base
than what Base imposes on itself already.
(good idea to file an enhancement report).
For that, I was trying earlier today to find the exact instances
of when there is a warning, and when there is not. I didn't get
warnings to come up consistently. (Even the case I described in
question 2 doens't always give a warning.) I'd have to take a
look at this some time again.
-- Simon
