Re: [Boost-users] boost variant is not a literal type

Subject: Re: [Boost-users] boost variant is not a literal type
From: Agustín K-ballo Bergé (kaballo86_at_[hidden])
Date: 2016-01-06 16:03:02

On 1/6/2016 5:52 PM, Robert Ramey wrote:
> I want to use Boost.Variant constexpr function. In order to do this it
> has to be a literal type - which apparently it isn't.  The rule for
> being a literal type are summarized here:
>
> http://en.cppreference.com/w/cpp/concept/LiteralType
>
> a boost::variant isn't default constructable so it fails to qualify.

A default constructor is not a requirement of literal type.

> Has anyone else had this problem before and managed to solve it in a
> convenient way?

Constexpr support is a complicated thing for `variant`, and it would 
require a complete reimplementation of `boost::variant`. The trickiest 
part is having a trivial destructor. For the gory details have a look at 
these articles:

https://akrzemi1.wordpress.com/2012/12/13/constexpr-unions/

http://talesofcpp.fusionfenix.com/post-20/eggs.variant---part-ii-the-constexpr-experience

Also note that literal unions are over restricted, and that restricts 
the kind of member types a literal variant might have. That's CWG2096 
http://open-std.org/jtc1/sc22/wg21/docs/cwg_active.html#2096

Regards,
-- 
Agustín K-ballo Bergé.-
http://talesofcpp.fusionfenix.com