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Construct a variant using the value #9

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douglasantos00 opened this issue Jul 1, 2016 · 1 comment
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Construct a variant using the value #9

douglasantos00 opened this issue Jul 1, 2016 · 1 comment
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@douglasantos00
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Right now, to construct a variant we need to create an empty one and them use set or emplace. Why not having a constructor that takes the value directly? We already have it for operator=:

<typename U>
  U &operator =(U const &value)
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juchem commented Jul 1, 2016

IIRC we didn't reach a good enough API regarding overload resolution.
Sounds reasonable, though. I'll take a stab at it again.

@juchem juchem self-assigned this Sep 7, 2016
@yfeldblum yfeldblum closed this as not planned Won't fix, can't repro, duplicate, stale Jun 6, 2024
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@yfeldblum @juchem @douglasantos00