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Subject: Re: [Boost-users] boost serialization + boost variant + boost blank
From: Andreas M. Iwanowski (namezero_at_[hidden])
Date: 2015-02-01 15:25:10

If you must serialize the variant then a serialize() will have to be defined for every type.

Unless I have misunderstood what you mean by "really works":
It's not about satisfying the compiler, but rather telling it how to serialize this member.
By telling it (with an empty serialize() ) to not serialize or de-serialize anything, it works appropriately, as long as you read/write the same data in the same order (i.e. nothing :) )

Mit freundlichen Grüßen / With best regards

Andreas Iwanowski - IT Administrator / Software Developer | namezero_at_[hidden]
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-----Original Message-----
From: Boost-users [mailto:boost-users-bounces_at_[hidden]] On Behalf Of Merrill Cornish
Sent: Sunday, 01 February, 2015 21:16
To: boost-users_at_[hidden]
Subject: Re: [Boost-users] boost serialization + boost variant + boost blank

If simply defining an empty serialize() function really works, then your suggestion is best. Next, question, does it really work--or just temporarily satisfy the compiler.
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