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# Move constructors

From cppreference.com
< cpp‎ | language

A move constructor of class `T` is a non-template constructor whose first parameter is T&&, const T&&, volatile T&&, or const volatile T&&, and either there are no other parameters, or the rest of the parameters all have default values.

### [edit] Syntax

class_name ( class_name `&&` ) | (1) | (since C++11)
class_name ( class_name `&&` ) = default; | (2) | (since C++11)
class_name ( class_name `&&` ) = delete; | (3) | (since C++11)

### [edit] Explanation

1. Typical declaration of a move constructor.
2. Forcing a move constructor to be generated by the compiler.
3. Avoiding implicit move constructor.

The move constructor is called whenever selected by overload resolution, which typically occurs when an object is initialized (by direct-initialization or copy-initialization) from rvalue (xvalue or prvalue) (until C++17)xvalue (since C++17) of the same type, including

- initialization: T a = std::move(b); or T a(std::move(b));, where b is of type `T`;
- function argument passing: f(std::move(a));, where `a` is of type `T` and `f` is void f(T t);
- function return: return a; inside a function such as T f(), where `a` is of type `T` which has a move constructor.

Move constructors typically "steal" the resources held by the argument (e.g. pointers to dynamically-allocated objects, file descriptors, TCP sockets, I/O streams, running threads, etc), rather than make copies of them, and leave the argument in some valid but otherwise indeterminate state. For example, moving from a std::string or from a std::vector may result in the argument being left empty. However, this behavior should not be relied upon. For some types, such as std::unique_ptr, the moved-from state is fully specified.

### [edit] 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, | (until C++14)

then the compiler will declare a move constructor as a non-explicit `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`.

The implicitly-declared (or defaulted on its first declaration) move constructor has an exception specification as described in dynamic exception specification (until C++17)exception specification (since C++17)

### [edit] Deleted implicitly-declared move constructor

The implicitly-declared or defaulted move constructor for class `T` is defined as deleted in any of the following is true:

- `T` has non-static data members that cannot be moved (have deleted, inaccessible, or ambiguous move constructors);
- `T` has direct or virtual base class that cannot be moved (has deleted, inaccessible, or ambiguous move constructors);
- `T` has direct or virtual base class with a deleted or inaccessible destructor;
- `T` is a union and has a variant member with non-trivial copy constructor;

`T` has a non-static data member or a direct or virtual base without a move constructor that is not trivially copyable. | (until C++14)

The deleted implicitly-declared move constructor is ignored by overload resolution (otherwise it would prevent copy-initialization from rvalue). | (since C++14)

### [edit] Trivial move constructor

The move constructor for class `T` is trivial if all of the following is true:

- it is not user-provided (meaning, it is implicitly-defined or defaulted);
- `T` has no virtual member functions;
- `T` has no virtual base classes;
- the move constructor selected for every direct base of `T` is trivial;
- the move constructor selected for every non-static class type (or array of class type) member of `T` is trivial;

`T` has no non-static data members of volatile-qualified type. | (since C++14)

A trivial move constructor is a constructor that performs the same action as the trivial copy constructor, that is, makes a copy of the object representation as if by std::memmove. All data types compatible with the C language (POD types) are trivially movable.

### [edit] Implicitly-defined move constructor

If the implicitly-declared move constructor is neither deleted nor trivial, it is defined (that is, a function body is generated and compiled) by the compiler if odr-used. For union types, the implicitly-defined move constructor copies the object representation (as by std::memmove). For non-union class types (class and struct), the move constructor performs full member-wise move of the object's bases and non-static members, in their initialization order, using direct initialization with an xvalue argument.

### [edit] Notes

To make strong exception guarantee possible, user-defined move constructors should not throw exceptions. For example, std::vector relies on std::move_if_noexcept to choose between move and copy when the elements need to be relocated.

If both copy and move constructors are provided, overload resolution selects the move constructor if the argument is an rvalue (either a prvalue such as a nameless temporary or an xvalue such as the result of std::move), and selects the copy constructor if the argument is an lvalue (named object or a function/operator returning lvalue reference). If only the copy constructor is provided, all argument categories select it (as long as it takes a reference to const, since rvalues can bind to const references), which makes copying the fallback for moving, when moving is unavailable.

In many situations, move constructors are optimized out even if they would produce observable side-effects, see copy elision.

A constructor is called a 'move constructor' when it takes an rvalue reference as a parameter. It is not obligated to move anything, the class is not required to have a resource to be moved and a 'move constructor' may not be able to move a resource as in the allowable (but maybe not sensible) case where the parameter is a const rvalue reference (const T&&).

### [edit] Example

Run this code

    #include <string>
    #include <iostream>
    #include <iomanip>
    #include <utility>

    struct A
    {
        std::string s;
        A() : s("test") { }
        A(const A& o) : s(o.s) { std::cout << "move failed!\n"; }
        A(A&& o) noexcept : s(std::move(o.s)) { }
    };

    A f(A a)
    {
        return a;
    }

    struct B : A
    {
        std::string s2;
        int n;
        // implicit move constructor B::(B&&)
        // calls A's move constructor
        // calls s2's move constructor
        // and makes a bitwise copy of n
    };

    struct C : B
    {
        ~C() { } // destructor prevents implicit move ctor C::(C&&)
    };

    struct D : B
    {
        D() { }
        ~D() { }          // destructor would prevent implicit move ctor D::(D&&)
        D(D&&) = default; // force a move ctor anyway
    };

    int main()
    {
        std::cout << "Trying to move A\n";
        A a1 = f(A()); // move-construct from rvalue temporary
        A a2 = std::move(a1); // move-construct from xvalue

        std::cout << "Trying to move B\n";
        B b1;
        std::cout << "Before move, b1.s = " << std::quoted(b1.s) << "\n";
        B b2 = std::move(b1); // calls implicit move ctor
        std::cout << "After move, b1.s = " << std::quoted(b1.s) << "\n";

        std::cout << "Trying to move C\n";
        C c1;
        C c2 = std::move(c1); // calls the copy constructor

        std::cout << "Trying to move D\n";
        D d1;
        D d2 = std::move(d1);
    }

Output:

    Trying to move A
    Trying to move B
    Before move, b1.s = "test"
    After move, b1.s = ""
    Trying to move C
    move failed!
    Trying to move D

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