Re: which compiler is right (either to compile or to barf)...

[LIU Hao <lh_mouse@126.com>]

[-- Attachment #1.1: Type: text/plain, Size: 2746 bytes --]

在 2022/12/2 15:18, leon zadorin via Gcc-help 写道:
> (1) On the one hand, given that I do instantiate ::std::optional with S
> (whith has member v whose elements' types, I guess(?) are at the line of
> declaring 'static ::std::optional<S> s', are still incompletely denifed,
> i.e. X) -- the clang discussion appears to say that at this moment optional
> is instantiated and S is incomplete (?) ... as so it is a UB... which I
> understood Jonathan's comment to relate to as well.
> 

Yes this requires `S` to be complete.


> (2) However, from your comment it would appear that 'instantiation of
> members of class template' is done afterwards anyways...
> 

Only `S` itself is required to be complete.


In your code:

   struct S {
       ::std::vector<::std::shared_ptr<J<X>>> v;
   };

`S` _is_ complete. It does not matter whether it contains a `vector` of or a `shared_ptr` to 
incomplete types, because the standard allows so. ([vector.overview] 4, [util.smartptr.shared] 2)


> I suppose to reconcile both bits of info:
> 
> (a) 'instantitation of members of class template' ... is it the same for
> instantitaion of the class template itself? (because UB specs appears to
> provide restrictions on std lib class instantiations, not specifically to
> inner members, but just for whole class)?; and
> 

No.

   ```
   struct T;
   extern T* eptr;  // okay; pointer to incomplete type

   struct S
     {
       shared_ptr<T> mptr;  // okay; `shared_ptr` to incomplete type
     };

   void
   use_s()
     {
       S s;  // oayk; `S` is complete

       s.mptr.reset();  // okay; `reset()` does not require `T` to
                        // be complete

       s.mptr.reset(::eptr);  // error; instantiates the default
                              // deleter expression which requires
                              // `T` to be complete
     }
   ```


> (b) in the above code right at the line 'static ::std::optional<S> s' is S
> considered incomplete type at that moment?
> 

No. See above.


> ... I'm just trying to figure out whether template "::std::optional<S>" is,
> itself, being instantiated at that very line (never mind its members) and
> if type S at that moment is considered to be incomplete... in which case it
> is UB... if not then may be its not a UB... I am confused :) ha ha :)
> wouldn't be the first time though :)

If `shared_ptr` is a bad example for you, here is a much simpler one:

   ```
   struct incomplete;

   struct complete
     {
       incomplete* ptr;  // okay
       incomplete& ref;  // also okay
       incomplete  val;  // invalid use of incomplete type
     };
   ```

-- 
Best regards,
LIU Hao


[-- Attachment #2: OpenPGP digital signature --]
[-- Type: application/pgp-signature, Size: 840 bytes --]