How to enforce move semantics when a vector grows?

Question

I have a std::vector of objects of a certain class A. The class is non-trivial and has copy constructors and move constructors defined.

Answer 1

You need to inform C++ (specifically std::vector) that your move constructor and destructor does not throw, using noexcept. Then the move constructor will be called when the vector grows.

This is how to declare and implement a move constuctor that is respected by std::vector:

A(A && rhs) noexcept { 
  std::cout << "i am the move constr" <<std::endl;
  ... some code doing the move ...  
  m_value=std::move(rhs.m_value) ; // etc...
}

If the constructor is not noexcept, std::vector can't use it, since then it can't ensure the exception guarantees demanded by the standard.

For more about what's said in the standard, read C++ Move semantics and Exceptions

Credit to Bo who hinted that it may have to do with exceptions. Also consider Kerrek SB's advice and use emplace_back when possible. It can be faster (but often is not), it can be clearer and more compact, but there are also some pitfalls (especially with non-explicit constructors).

Edit, often the default is what you want: move everything that can be moved, copy the rest. To explicitly ask for that, write

A(A && rhs) = default;

Doing that, you will get noexcept when possible: Is the default Move constructor defined as noexcept?

Note that early versions of Visual Studio 2015 and older did not support that, even though it supports move semantics.

Answer 2

Interestingly, gcc 4.7.2's vector only uses move constructor if both the move constructor and the destructor are noexcept. A simple example:

struct foo {
    foo() {}
    foo( const foo & ) noexcept { std::cout << "copy\n"; }
    foo( foo && ) noexcept { std::cout << "move\n"; }
    ~foo() noexcept {}
};

int main() {
    std::vector< foo > v;
    for ( int i = 0; i < 3; ++i ) v.emplace_back();
}

This outputs the expected:

move
move
move

However, when I remove noexcept from ~foo(), the result is different:

copy
copy
copy

I guess this also answers this question.

Answer 3

It seems, that the only way (for C++17 and early), to enforce std::vector use move semantics on reallocation is deleting copy constructor :) . In this way it will use your move constructors or die trying, at compile time :).

There are many rules where std::vector MUST NOT use move constructor on reallocation, but nothing about where it MUST USE it.

template<class T>
class move_only : public T{
public:
   move_only(){}
   move_only(const move_only&) = delete;
   move_only(move_only&&) noexcept {};
   ~move_only() noexcept {};

   using T::T;   
};

Live

or

template<class T>
struct move_only{
   T value;

   template<class Arg, class ...Args, typename = std::enable_if_t<
            !std::is_same_v<move_only<T>&&, Arg >
            && !std::is_same_v<const move_only<T>&, Arg >
    >>
   move_only(Arg&& arg, Args&&... args)
      :value(std::forward<Arg>(arg), std::forward<Args>(args)...)
   {}

   move_only(){}
   move_only(const move_only&) = delete;   
   move_only(move_only&& other) noexcept : value(std::move(other.value)) {};    
   ~move_only() noexcept {};   
};

Live code

Your T class must have noexcept move constructor/assigment operator and noexcept destructor. Otherwise you'll get compilation error.

std::vector<move_only<MyClass>> vec;