C++17 presents std::variant
and std::any
, both able to store different type of values under an object. For me, they are somehow similar (are they?).
Also std::variant
restricts the entry types, beside this one. Why we should prefer std::variant
over std::any
which is simpler to use?
The more things you check at compile time the fewer runtime bugs you have.
variant
guarantees that it contains one of a list of types (plus valueless by exception). It provides a way for you to guarantee that code operating on it considers every case in the variant with std::visit
; even every case for a pair of variant
s (or more).
any
does not. With any
the best you can do is "if the type isn't exactly what I ask for, some code won't run".
variant
exists in automatic storage. any
may use the free store; this means any
has performance and noexcept(false)
issues that variant
does not.
Checking for which of N types is in it is O(N) for an any
-- for variant
it is O(1).
any
is a dressed-up void*
. variant
is a dressed-up union
.
any
cannot store non-copy or non-move able types. variant
can.
The type of variant
is documentation for the reader of your code.
Passing a variant<Msg1, Msg2, Msg3>
through an API makes the operation obvious; passing an any
there means understanding the API requires reliable documentation or reading the implementation source.
Anyone who has been frustrated by statically typeless languages will understand the dangers of any
.
Now this doesn't mean any
is bad; it just doesn't solve the same problems as variant
. As a copyable object for type erasure purposes, it can be great. Runtime dynamic typing has its place; but that place is not "everywhere" but rather "where you cannot avoid it".