What are the differences between std::variant and boost::variant?

问题

In an answer to this SO question:

What is the equivalent of boost::variant in the C++ standard library?

it is mentioned that boost::variant and std::variant differ somewhat.

• What are the differences, as far as someone using these classes is concerned?
• What motivation did the committee express to adopt std::variant with these differences?
• What should I watch out for when coding with either of these, to maintain maximum compatibility with switching to the other one?

(the motivation is using boost::variant in pre-C++17 code)

回答1:

• Assignment/emplacement behavior:

  • boost::variant may allocate memory when performing assignment into a live variant. There are a number of rules that govern when this can happen, so whether a boost::variant will allocate memory depends on the Ts it is instantiated with.

  • std::variant will never dynamically allocate memory. However, as a concession to the complex rules of C++ objects, if an assignment/emplacement throws, then the variant may enter the "valueless_by_exception" state. In this state, the variant cannot be visited, nor will any of the other functions for accessing a specific member work.

    You can only enter this state if assignment/emplacement throws.

• Boost.Variant includes recursive_variant, which allows a variant to contain itself. They're essentially special wrappers around a pointer to a boost::variant, but they are tied into the visitation machinery.

  std::variant has no such helper type.

• std::variant offers more use of post-C++11 features. For example:

  • It forwards the noexcept status of the special member functions of its constituent types.

  • It has variadic template-based in-place constructors and emplacement functions.

  • Defect resolutions applied to C++17 may mean that it will also forward trivial copyability of its types. That is, if all of the types are trivially copyable, then so too will variant<Ts>.

回答2:

It seems the main point of contention regarding the design of a variant class has been what should happen when an assignment to the variant, which should upon completion destory the old value, throws an exception:

variant<std::string, MyClassWithThrowingDefaultCtor> v = "ABC";
v = MyClassWithThrowingDefaultCtor();

The options seem to be:

• Prevent this by restricting the possible representable types to nothrow-move-constructible ones.
• Keep the old value - but this requires double-buffers (which is what boost::variant does apparently).
• Have a 'disengaged' state with no value for each variant, and go to that state on such failures.
• Undefined behavior
• Make the variant throw when trying to read its value after something like that happens

and if I'm not mistaken, the latter is what's been accepted.

_{This is summarized from the ISO C++ blog post by Axel Naumann from Nov 2015.}

来源：https://stackoverflow.com/questions/40201371/what-are-the-differences-between-stdvariant-and-boostvariant