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Assignment operators

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Assignment operators modify the value of the object.

Operator name Syntax Over​load​able Prototype examples (for class T)
Inside class definition Outside class definition
simple assignment a = b Yes T& T::operator =(const T2& b); N/A
addition assignment a += b Yes T& T::operator +=(const T2& b); T& operator +=(T& a, const T2& b);
subtraction assignment a -= b Yes T& T::operator -=(const T2& b); T& operator -=(T& a, const T2& b);
multiplication assignment a *= b Yes T& T::operator *=(const T2& b); T& operator *=(T& a, const T2& b);
division assignment a /= b Yes T& T::operator /=(const T2& b); T& operator /=(T& a, const T2& b);
modulo assignment a %= b Yes T& T::operator %=(const T2& b); T& operator %=(T& a, const T2& b);
bitwise AND assignment a &= b Yes T& T::operator &=(const T2& b); T& operator &=(T& a, const T2& b);
bitwise OR assignment a |= b Yes T& T::operator |=(const T2& b); T& operator |=(T& a, const T2& b);
bitwise XOR assignment a ^= b Yes T& T::operator ^=(const T2& b); T& operator ^=(T& a, const T2& b);
bitwise left shift assignment a <<= b Yes T& T::operator <<=(const T2& b); T& operator <<=(T& a, const T2& b);
bitwise right shift assignment a >>= b Yes T& T::operator >>=(const T2& b); T& operator >>=(T& a, const T2& b);
Notes
  • All built-in assignment operators return *this, and most user-defined overloads also return *this so that the user-defined operators can be used in the same manner as the built-ins. However, in a user-defined operator overload, any type can be used as return type (including void).
  • T2 can be any type including T

Contents

[edit] Explanation

copy assignment operator replaces the contents of the object a with a copy of the contents of b (b is not modified). For class types, this is a special member function, described in copy assignment operator.

move assignment operator replaces the contents of the object a with the contents of b, avoiding copying if possible (b may be modified). For class types, this is a special member function, described in move assignment operator. (since C++11)

For non-class types, copy and move assignment are indistinguishable and are referred to as direct assignment.

compound assignment operators replace the contents the contents of the object a with the result of a binary operation between the previous value of a and the value of b.

[edit] Builtin direct assignment

For every type T, the following function signatures participate in overload resolution:

T*& operator=(T*&, T*);
T*volatile & operator=(T*volatile &, T*);

For every enumeration or pointer to member type T, optionally volatile-qualified, the following function signature participates in overload resolution:

T& operator=(T&, T );

For every pair A1 and A2, where A1 is an arithmetic type (optionally volatile-qualified) and A2 is a promoted arithmetic type, the following function signature participates in overload resolution:

A1& operator=(A1&, A2);

For expressions E1 of any scalar type T, the following additional forms of the builtin assignment expression are allowed:

E1 = {}
(since C++11)
E1 = {E2}
(since C++11)

Note: the above includes all non-class types except reference types, array types, function types, and the type void, which are not directly assignable.

The direct assignment operator expects a modifiable lvalue as its left operand and an rvalue expression or a braced-init-list (since C++11) as its right operand, and returns an lvalue identifying the left operand after modification.

For non-class types, the right operand is first implicitly converted to the cv-unqualified type of the left operand, and then its value is copied into the object identified by left operand.

When the left operand has reference type, the assignment operator modifies the referred-to object.

If the left and the right operands identify overlapping objects, the behavior is undefined (unless the overlap is exact and the type is the same)

If the right operand is a braced-init-list

  • the expression E1 = {} is equivalent to E1 = T{}, where T is the type of E1.
  • the expression E1 = {E2} is equivalent to E1 = T{E2}, where T is the type of E1.

For class types, the syntax E = {e1, e2, e3} generates a call to the assignment operator with the braced-init-list as the argument, which then selects the appropriate assignment operator following the rules of overload resolution

(since C++11)

[edit] Example

#include <iostream>
int main()
{
    int n = 0;  // not an assignment
    n = 1;      // direct assignment
    std::cout << n << ' ';
    n = {};     // zero-initialization, then assignment
    std::cout << n << ' ';
    n = 'a';    // integral promotion, then assignment
    std::cout << n << ' ';
    n = {'b'};   // explicit cast, then assignment
    std::cout << n << ' ';
    n = 1.0;    // floating-point conversion, then assignment
    std::cout << n << ' ';
//    n = {1.0}; // compiler error (narrowing conversion)
 
    int& r = n;  // not an assignment
    int* p;
 
    r = 2;       // assignment through reference
    std::cout << n << '\n';
    p = &n;      // direct assignment
    p = nullptr; // null-pointer conversion, then assignment 
 
    struct {int a; std::string s;} obj;
    obj = {1, "abc"}; // assignment from a braced-init-list
    std::cout << obj.a << ':' << obj.s << '\n';
}

Output:

1 0 97 98 1 2
1:abc

[edit] Builtin compound assignment

For every pair A1 and A2, where A1 is an arithmetic type (optionally volatile-qualified) and A2 is a promoted arithmetic type, the following function signatures participate in overload resolution:

A1& operator*=(A1&, A2);
A1& operator/=(A1&, A2);
A1& operator+=(A1&, A2);
A1& operator-=(A1&, A2);

For every pair I1 and I2, where I1 is an integral type (optionally volatile-qualified) and I2 is a promoted integral type, the following function signatures participate in overload resolution:

I1& operator%=(I1&, I2);
I1& operator<<=(I1&, I2);
I1& operator>>=(I1&, I2);
I1& operator&=(I1&, I2);
I1& operator^=(I1&, I2);
I1& operator|=(I1&, I2);

For every optionally cv-qualified object type T, the following function signatures participate in overload resolution:

T*& operator+=(T*&, std::ptrdiff_t);
T*& operator-=(T*&, std::ptrdiff_t);
T*volatile & operator+=(T*volatile &, std::ptrdiff_t);
T*volatile & operator-=(T*volatile &, std::ptrdiff_t);

The behavior of every builtin compound-assignment expression E1 op= E2 (where E1 is a modifiable lvalue expression and E2 is an rvalue expression or a braced-init-list (since C++11)) is exactly the same as the behavior of the expression E1 = E1 op E2, except that the expression E1 is evaluated only once and that it behaves as a single operation with respect to indeterminately-sequenced function calls (e.g. in f(a += b, g()), the += is either not started at all or is completed as seen from inside g()).

[edit] Example

[edit] See also

Operator precedence

Common operators
assignment increment
decrement
arithmetic logical comparison member
access
other

a = b
a += b
a -= b
a *= b
a /= b
a %= b
a &= b
a |= b
a ^= b
a <<= b
a >>= b

++a
--a
a++
a--

+a
-a
a + b
a - b
a * b
a / b
a % b
~a
a & b
a | b
a ^ b
a << b
a >> b

!a
a && b
a || b

a == b
a != b
a < b
a > b
a <= b
a >= b

a[b]
*a
&a
a->b
a.b
a->*b
a.*b

a(...)
a, b
(type) a
? :

Special operators

static_cast converts one type to another compatible type
dynamic_cast converts virtual base class to derived class
const_cast converts type to compatible type with different cv qualifiers
reinterpret_cast converts type to incompatible type
new allocates memory
delete deallocates memory
sizeof queries the size of a type
sizeof... queries the size of a parameter pack (since C++11)
typeid queries the type information of a type
noexcept checks if an expression can throw an exception (since C++11)
alignof queries alignment requirements of a type (since C++11)