It seems that a vector will check if the move constructor is labeled as noexcept before deciding on whether to move or copy elements when reallocating. Is the default move constructor defined as noexcept? I saw the following documentation but it didn't specify this.http://en.cppreference.com/w/cpp/language/move_constructor
Implicitly-declared move constructor
If no user-defined move constructors are provided for a class type (struct, class, or union), and all of the following is true: there are no user-declared copy constructors there are no user-declared copy assignment operators there are no user-declared move assignment operators there are no user-declared destructors the implicitly-declared move constructor is not defined as deleted due to conditions detailed in the next section then the compiler will declare a move constructor as an inline public member of its class with the signature T::T(T&&) A class can have multiple move constructors, e.g. both T::T(const T&&) and T::T(T&&). If some user-defined move constructors are present, the user may still force the generation of the implicitly declared move constructor with the keyword default.
I think the answer is 15.4/14 (Exception specifications):
An inheriting constructor (12.9) and an implicitly declared special member function (Clause 12) have an exception-specification. If
f
is an inheriting constructor or an implicitly declared default constructor, copy constructor, move constructor, destructor, copy assignment operator, or move assignment operator, its implicit exception-specification specifies the type-idT
if and only ifT
is allowed by the exception-specification of a function directly invoked byf
’s implicit definition;f
allows all exceptions if any function it directly invokes allows all exceptions, andf
has the exception-specificationnoexcept(true)
if every function it directly invokes allows no exceptions.
Basically, it Does What You Think, and the implicitly-declared move constructor is noexcept
whenever it can be.