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boost::variant seems a powerful container to manipulate a heterogeneous set of types. I am wondering its cost. In memory, I think it takes up the size of the largest type plus an integer representing which(). For apply_visitor(), I think its performance is very good, it can call directly the function other than lots of ifs. Are my points right?
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Boost. Variant provides a class called boost::variant that resembles union . You can store values of different types in a boost::variant variable. At any point only one value can be stored.
Its purpose is to demonstrate that std::variant can use one of the variant type's converting constructors as long as it is unambiguous. std::variant<std::string, int> x("abc"); instead, to show that off more clearly.
(since C++17) The class template std::variant represents a type-safe union. An instance of std::variant at any given time either holds a value of one of its alternative types, or in the case of error - no value (this state is hard to achieve, see valueless_by_exception).
You're almost right.
The size of boost::variant is is the max size of any element, rounded up as needed for the largest alignment, plus the size of some integer, and again rounded up.
Think about a variant of these types, assuming the tag is uint32_t:
struct foo { uint32_t value[3]; }; // size 12, align 4
struct bar { uint64_t v2; }; // size 8, align 8
An untagged union must have size 16, align 8; adding the 4-byte tag must go up to size 24 to keep align 8.
Or consider a variant of:
struct foo { uint8_t value[5]; }; // size 5, align 1
struct bar { uint16_t v2; }; // size 2, align 2
An untagged union of these must have size 6, align 2; adding the 4-byte tag forces you to size 12, align 4.
For calling, I expect it uses an array-of-functions lookup (that's how I implemented my own variant, which was necessary since boost's didn't support move constructors), since if chains don't perform well and switches are impossible.
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