Call Default Assignment Operator Definition

A move assignment operator of class is a non-template non-static member function with the name operator= that takes exactly one parameter of type T&&, const T&&, volatile T&&, or constvolatile T&&.

[edit]Syntax

class_nameclass_name ( class_name ) (1) (since C++11)
class_nameclass_name ( class_name ) = default; (2) (since C++11)
class_nameclass_name ( class_name ) = delete; (3) (since C++11)

[edit]Explanation

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

The move assignment operator is called whenever it is selected by overload resolution, e.g. when an object appears on the left-hand side of an assignment expression, where the right-hand side is an rvalue of the same or implicitly convertible type.

Move assignment operators 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, move-assigning from a std::string or from a std::vector may result in the argument being left empty. This is not, however, a guarantee. A move assignment is less, not more restrictively defined than ordinary assignment; where ordinary assignment must leave two copies of data at completion, move assignment is required to leave only one.

[edit]Implicitly-declared move assignment operator

If no user-defined move assignment operators 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 move constructors;
  • there are no user-declared copy assignment operators;
  • there are no user-declared destructors;
  • the implicitly-declared move assignment operator would not be defined as deleted,
(until C++14)

then the compiler will declare a move assignment operator as an member of its class with the signature .

A class can have multiple move assignment operators, e.g. both T& T::operator=(const T&&) and T& T::operator=(T&&). If some user-defined move assignment operators are present, the user may still force the generation of the implicitly declared move assignment operator with the keyword .

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

Because some assignment operator (move or copy) is always declared for any class, the base class assignment operator is always hidden. If a using-declaration is used to bring in the assignment operator from the base class, and its argument type could be the same as the argument type of the implicit assignment operator of the derived class, the using-declaration is also hidden by the implicit declaration.

[edit]Deleted implicitly-declared move assignment operator

The implicitly-declared or defaulted move assignment operator for class is defined as deleted if any of the following is true:

  • has a non-static data member that is const;
  • has a non-static data member of a reference type;
  • has a non-static data member that cannot be move-assigned (has deleted, inaccessible, or ambiguous move assignment operator);
  • has direct or virtual base class that cannot be move-assigned (has deleted, inaccessible, or ambiguous move assignment operator);
  • has a non-static data member or a direct or virtual base without a move assignment operator that is not trivially copyable;
  • has a direct or indirect virtual base class.
(until C++14)

A deleted implicitly-declared move assignment operator is ignored by overload resolution.

(since C++14)

[edit]Trivial move assignment operator

The move assignment operator for class is trivial if all of the following is true:

  • It is not user-provided (meaning, it is implicitly-defined or defaulted);
  • has no virtual member functions;
  • has no virtual base classes;
  • the move assignment operator selected for every direct base of is trivial;
  • the move assignment operator selected for every non-static class type (or array of class type) member of is trivial;
  • has no non-static data members of volatile-qualified type.
(since C++14)

A trivial move assignment operator performs the same action as the trivial copy assignment operator, 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 move-assignable.

[edit]Implicitly-defined move assignment operator

If the implicitly-declared move assignment operator 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 assignment operator copies the object representation (as by std::memmove).

For non-union class types (class and struct), the move assignment operator performs full member-wise move assignment of the object's direct bases and immediate non-static members, in their declaration order, using built-in assignment for the scalars, memberwise move-assignment for arrays, and move assignment operator for class types (called non-virtually).

As with copy assignment, it is unspecified whether virtual base class subobjects that are accessible through more than one path in the inheritance lattice, are assigned more than once by the implicitly-defined move assignment operator:

struct V { V& operator=(V&& other){// this may be called once or twice// if called twice, 'other' is the just-moved-from V subobjectreturn*this;}};struct A :virtual V {};// operator= calls V::operator=struct B :virtual V {};// operator= calls V::operator=struct C : B, A {};// operator= calls B::operator=, then A::operator=// but they may only called V::operator= once   int main(){ C c1, c2; c2 = std::move(c1);}
(since C++14)

[edit]Notes

If both copy and move assignment operators are provided, overload resolution selects the move assignment 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 assignment if the argument is an lvalue (named object or a function/operator returning lvalue reference). If only the copy assignment is provided, all argument categories select it (as long as it takes its argument by value or as reference to const, since rvalues can bind to const references), which makes copy assignment the fallback for move assignment, when move is unavailable.

It is unspecified whether virtual base class subobjects that are accessible through more than one path in the inheritance lattice, are assigned more than once by the implicitly-defined move assignment operator (same applies to copy assignment).

See assignment operator overloading for additional detail on the expected behavior of a user-defined move-assignment operator.

[edit]Example

Run this code

Output:

#include <string>#include <iostream>#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): s(std::move(o.s)){} A& operator=(const A& other){ s = other.s;std::cout<<"copy assigned\n";return*this;} A& operator=(A&& other){ s = std::move(other.s);std::cout<<"move assigned\n";return*this;}};   A f(A a){return a;}   struct B : A {std::string s2;int n;// implicit move assignment operator B& B::operator=(B&&)// calls A's move assignment operator// calls s2's move assignment operator// and makes a bitwise copy of n};   struct C : B { ~C(){}// destructor prevents implicit move assignment};   struct D : B { D(){} ~D(){}// destructor would prevent implicit move assignment D& operator=(D&&)=default;// force a move assignment anyway };   int main(){ A a1, a2;std::cout<<"Trying to move-assign A from rvalue temporary\n"; a1 = f(A());// move-assignment from rvalue temporarystd::cout<<"Trying to move-assign A from xvalue\n"; a2 = std::move(a1);// move-assignment from xvalue   std::cout<<"Trying to move-assign B\n"; B b1, b2;std::cout<<"Before move, b1.s = \""<< b1.s<<"\"\n"; b2 = std::move(b1);// calls implicit move assignmentstd::cout<<"After move, b1.s = \""<< b1.s<<"\"\n";   std::cout<<"Trying to move-assign C\n"; C c1, c2; c2 = std::move(c1);// calls the copy assignment operator   std::cout<<"Trying to move-assign D\n"; D d1, d2; d2 = std::move(d1);}
Trying to move-assign A from rvalue temporary move assigned Trying to move-assign A from xvalue move assigned Trying to move-assign B Before move, b1.s = "test" move assigned After move, b1.s = "" Trying to move-assign C copy assigned Trying to move-assign D move assigned

Assignment Operators

What is “self assignment”?

Self assignment is when someone assigns an object to itself. For example,

Obviously no one ever explicitly does a self assignment like the above, but since more than one pointer or reference can point to the same object (aliasing), it is possible to have self assignment without knowing it:

This is only valid for copy assignment. Self-assignment is not valid for move assignment.

Why should I worry about “self assignment”?

If you don’t worry about self assignment, you’ll expose your users to some very subtle bugs that have very subtle and often disastrous symptoms. For example, the following class will cause a complete disaster in the case of self-assignment:

If someone assigns a object to itself, line #1 deletes both and since and are the same object. But line #2 uses , which is no longer a valid object. This will likely cause a major disaster.

The bottom line is that you the author of class are responsible to make sure self-assignment on a object is innocuous. Do not assume that users won’t ever do that to your objects. It is your fault if your object crashes when it gets a self-assignment.

Aside: the above has a second problem: If an exception is thrown while evaluating (e.g., an out-of-memory exception or an exception in ’s copy constructor), will be a dangling pointer — it will point to memory that is no longer valid. This can be solved by allocating the new objects before deleting the old objects.

This is only valid for copy assignment. Self-assignment is not valid for move assignment.

Okay, okay, already; I’ll handle self-assignment. How do I do it?

You should worry about self assignment every time you create a class. This does not mean that you need to add extra code to all your classes: as long as your objects gracefully handle self assignment, it doesn’t matter whether you had to add extra code or not.

We will illustrate the two cases using the assignment operator in the previous FAQ:

  1. If self-assignment can be handled without any extra code, don’t add any extra code. But do add a comment so others will know that your assignment operator gracefully handles self-assignment:

    Example 1a:

    Example 1b:

  2. If you need to add extra code to your assignment operator, here’s a simple and effective technique:

    Or equivalently:

By the way: the goal is not to make self-assignment fast. If you don’t need to explicitly test for self-assignment, for example, if your code works correctly (even if slowly) in the case of self-assignment, then do not put an test in your assignment operator just to make the self-assignment case fast. The reason is simple: self-assignment is almost always rare, so it merely needs to be correct - it does not need to be efficient. Adding the unnecessary statement would make a rare case faster by adding an extra conditional-branch to the normal case, punishing the many to benefit the few.

In this case, however, you should add a comment at the top of your assignment operator indicating that the rest of the code makes self-assignment is benign, and that is why you didn’t explicitly test for it. That way future maintainers will know to make sure self-assignment stays benign, or if not, they will need to add the test.

This is only valid for copy assignment. Self-assignment is not valid for move assignment.

I’m creating a derived class; should my assignment operators call my base class’s assignment operators?

Yes (if you need to define assignment operators in the first place).

If you define your own assignment operators, the compiler will not automatically call your base class’s assignment operators for you. Unless your base class’s assignment operators themselves are broken, you should call them explicitly from your derived class’s assignment operators (again, assuming you create them in the first place).

However if you do not create your own assignment operators, the ones that the compiler create for you will automatically call your base class’s assignment operators.

Example:

0 Replies to “Call Default Assignment Operator Definition”

Lascia un Commento

L'indirizzo email non verrĂ  pubblicato. I campi obbligatori sono contrassegnati *