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Subject: [Boost-users] boost variant is not a literal type
From: Robert Ramey (ramey_at_[hidden])
Date: 2016-01-06 15:52:28

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:

a boost::variant isn't default constructable so it fails to qualify.

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

Robert Ramey

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