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# Copy assignment operator

From cppreference.com
< cpp‎ | language

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

### [edit] Syntax

class_name `&` class_name `:: operator=` ( class_name ) | (1) |
class_name `&` class_name `:: operator=` ( const class_name `&` ) | (2) |
class_name `&` class_name `:: operator=` ( const class_name `&` ) = default; | (3) | (since C++11)
class_name `&` class_name `:: operator=` ( const class_name `&` ) = delete; | (4) | (since C++11)

### [edit] Explanation

1. Typical declaration of a copy assignment operator when copy-and-swap idiom can be used
2. Typical declaration of a copy assignment operator when copy-and-swap idiom cannot be used
3. Forcing a copy assignment operator to be generated by the compiler
4. Avoiding implicit copy assignment

The copy assignment operator is called whenever selected by overload resolution, e.g. when an object appears on the left side of an assignment expression.

### [edit] Implicitly-declared copy assignment operator

If no user-defined copy assignment operators are provided for a class type (struct, class, or union), the compiler will always declare one as an inline public member of the class. This implicitly-declared copy assignment operator has the form T& T::operator=(const T&) if all of the following is true:

- each direct base `B` of `T` has a copy assignment operator whose parameters are `B` or `const B&` or const volatile B&
- each non-static data member `M` of `T` of class type or array of class type has a copy assignment operator whose parameters are `M` or `const M&` or const volatile M&

Otherwise the implicitly-declared copy assignment operator is declared as T& T::operator=(T&). (Note that due to these rules, the implicitly-declared copy assignment operator cannot bind to a volatile lvalue argument)

A class can have multiple copy assignment operators, e.g. both T& T::operator=(const T&) and T& T::operator=(T). If some user-defined copy assignment operators are present, the user may still force the generation of the implicitly declared copy assignment operator with the keyword `default`. (since C++11)

Because the copy assignment operator 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 copy assignment operator

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

- `T` has a non-static data member that is const
- `T` has a non-static data member of a reference type.
- `T` has a non-static data member that cannot be copy-assigned (has deleted, inaccessible, or ambiguous copy assignment operator)
- `T` has direct or virtual base class that cannot be copy-assigned (has deleted, inaccessible, or ambiguous move assignment operator)
- `T` has a user-declared move constructor
- `T` has a user-declared move assignment operator

### [edit] Trivial copy assignment operator

The copy assignment operator for class `T` is trivial if all of the following is true:

- The operator is not user-provided (meaning, it is implicitly-defined or defaulted), and if it is defaulted, its signature is the same as implicitly-defined
- `T` has no virtual member functions
- `T` has no virtual base classes
- The copy assignment operator selected for every direct base of `T` is trivial
- The copy assignment operator selected for every non-static class type (or array of class type) memeber of `T` is trivial

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

A trivial copy assignment operator makes a copy of the object representation as if by std::memmove. All data types compatible with the C language (POD types) are trivially copy-assignable.

### [edit] Implicitly-defined copy assignment operator

If the implicitly-declared copy assignment operator is not deleted or trivial, it is defined (that is, a function body is generated and compiled) by the compiler. For union types, the implicitly-defined copy assignment copies the object representation (as by std::memmove). For non-union class types (class and struct), the operator performs member-wise copy assignment of the object's bases and non-static members, in their initialization order, using, using built-in assignment for the scalars and copy assignment operator for class types.

The generation of the implicitly-defined copy assignment operator is deprecated(since C++11) if `T` has a user-declared destructor or user-declared copy constructor.

### [edit] Notes

If both copy and move assignment operators are provided, overload resolution selects the move assignment if the argument is an rvalue (either prvalue such as a nameless temporary or xvalue such as the result of std::move), and selects the copy assignment if the argument is 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.

### [edit] Copy and swap

Copy assignment operator can be expressed in terms of copy constructor, destructor, and the swap() member function, if one is provided:

T& T::operator=(T arg) { // copy/move constructor is called to construct arg
swap(arg); // resources exchanged between *this and arg
return *this;
} // destructor is called to release the resources formerly held by *this

For non-throwing swap(), this form provides strong exception guarantee. For rvalue arguments, this form automatically invokes the move constructor, and is sometimes referred to as "unifying assignment operator" (as in, both copy and move).

### [edit] Example

Run this code

    #include <iostream>
    #include <memory>
    #include <string>
    #include <algorithm>

    struct A {
        int n;
        std::string s1;
        // user-defined copy assignment, copy-and-swap form
        A& operator=(A other) {
            std::cout << "copy assignment of A\n";
            std::swap(n, other.n);
            std::swap(s1, other.s1);
            return *this;
        }
    };

    struct B : A {
        std::string s2;
        // implicitly-defined copy assignment
    };

    struct C {
         std::unique_ptr<int[]> data;
         std::size_t size;
         // non-copy-and-swap assignment
         C& operator=(const C& other) {
             // check for self-assignment
             if(&other == this)
                 return *this;
             // reuse storage when possible
             if(size != other.size)
                 data.reset(new int[other.size]);
             std::copy(&other.data[0],
                       &other.data[0] + std::min(size, other.size),
                       &data[0]);
             return *this;
         }
         // note: copy-and-swap would always cause a reallocation
    };

    int main()
    {
        A a1, a2;
        std::cout << "a1 = a2 calls ";
        a1 = a2; // user-defined copy assignment

        B b1, b2;
        b2.s1 = "foo";
        b2.s2 = "bar";
        std::cout << "b1 = b2 calls ";
        b1 = b2; // implicitly-defined copy assignment
        std::cout << "b1.s1 = " << b1.s1 << " b1.s2 = " << b1.s2 <<  '\n';
    }

Output:

    a1 = a2 calls copy assignment of A
    b1 = b2 calls copy assignment of A
    b1.s1 = foo b1.s2 = bar

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