# Assignment operators

## Syntax

expression assignment-operator expression

assignment-operator: one of
`=``*=``/=``%=``+=``-=``<<=``>>=``&=``^=``|=`

## Remarks

Assignment operators store a value in the object specified by the left operand. There are two kinds of assignment operations:

• simple assignment, in which the value of the second operand is stored in the object specified by the first operand.

• compound assignment, in which an arithmetic, shift, or bitwise operation is performed before storing the result.

All assignment operators in the following table except the `=` operator are compound assignment operators.

### Assignment operators table

Operator Meaning
`=` Store the value of the second operand in the object specified by the first operand (simple assignment).
`*=` Multiply the value of the first operand by the value of the second operand; store the result in the object specified by the first operand.
`/=` Divide the value of the first operand by the value of the second operand; store the result in the object specified by the first operand.
`%=` Take modulus of the first operand specified by the value of the second operand; store the result in the object specified by the first operand.
`+=` Add the value of the second operand to the value of the first operand; store the result in the object specified by the first operand.
`-=` Subtract the value of the second operand from the value of the first operand; store the result in the object specified by the first operand.
`<<=` Shift the value of the first operand left the number of bits specified by the value of the second operand; store the result in the object specified by the first operand.
`>>=` Shift the value of the first operand right the number of bits specified by the value of the second operand; store the result in the object specified by the first operand.
`&=` Obtain the bitwise AND of the first and second operands; store the result in the object specified by the first operand.
`^=` Obtain the bitwise exclusive OR of the first and second operands; store the result in the object specified by the first operand.
`|=` Obtain the bitwise inclusive OR of the first and second operands; store the result in the object specified by the first operand.

### Operator keywords

Three of the compound assignment operators have keyword equivalents. They are:

Operator Equivalent
`&=` `and_eq`
`|=` `or_eq`
`^=` `xor_eq`

C++ specifies these operator keywords as alternative spellings for the compound assignment operators. In C, the alternative spellings are provided as macros in the <iso646.h> header. In C++, the alternative spellings are keywords; use of <iso646.h> or the C++ equivalent <ciso646> is deprecated. In Microsoft C++, the `/permissive-` or `/Za` compiler option is required to enable the alternative spelling.

## Example

``````// expre_Assignment_Operators.cpp
// compile with: /EHsc
// Demonstrate assignment operators
#include <iostream>
using namespace std;
int main() {
int a = 3, b = 6, c = 10, d = 0xAAAA, e = 0x5555;

a += b;      // a is 9
b %= a;      // b is 6
c >>= 1;      // c is 5
d |= e;      // Bitwise--d is 0xFFFF

cout  << "a = 3, b = 6, c = 10, d = 0xAAAA, e = 0x5555" << endl
<< "a += b yields " << a << endl
<< "b %= a yields " << b << endl
<< "c >>= 1 yields " << c << endl
<< "d |= e yields " << hex << d << endl;
}
``````

## Simple assignment

The simple assignment operator (`=`) causes the value of the second operand to be stored in the object specified by the first operand. If both objects are of arithmetic types, the right operand is converted to the type of the left, before storing the value.

Objects of `const` and `volatile` types can be assigned to l-values of types that are only `volatile`, or that aren't `const` or `volatile`.

Assignment to objects of class type (`struct`, `union`, and `class` types) is performed by a function named `operator=`. The default behavior of this operator function is to perform a bitwise copy; however, this behavior can be modified using overloaded operators. For more information, see Operator overloading. Class types can also have copy assignment and move assignment operators. For more information, see Copy constructors and copy assignment operators and Move constructors and move assignment operators.

An object of any unambiguously derived class from a given base class can be assigned to an object of the base class. The reverse isn't true because there's an implicit conversion from derived class to base class, but not from base class to derived class. For example:

``````// expre_SimpleAssignment.cpp
// compile with: /EHsc
#include <iostream>
using namespace std;
class ABase
{
public:
ABase() { cout << "constructing ABase\n"; }
};

{
public:
};

int main()
{
ABase aBase;

}
``````

Assignments to reference types behave as if the assignment were being made to the object to which the reference points.

For class-type objects, assignment is different from initialization. To illustrate how different assignment and initialization can be, consider the code

``````UserType1 A;
UserType2 B = A;
``````

The preceding code shows an initializer; it calls the constructor for `UserType2` that takes an argument of type `UserType1`. Given the code

``````UserType1 A;
UserType2 B;

B = A;
``````

the assignment statement

``````B = A;
``````

can have one of the following effects:

• Call the function `operator=` for `UserType2`, provided `operator=` is provided with a `UserType1` argument.

• Call the explicit conversion function `UserType1::operator UserType2`, if such a function exists.

• Call a constructor `UserType2::UserType2`, provided such a constructor exists, that takes a `UserType1` argument and copies the result.

## Compound assignment

The compound assignment operators are shown in the Assignment operators table. These operators have the form e1 op= e2, where e1 is a non-`const` modifiable l-value and e2 is:

• an arithmetic type

• a pointer, if op is `+` or `-`

The e1 op= e2 form behaves as e1 `=` e1 op e2, but e1 is evaluated only once.

Compound assignment to an enumerated type generates an error message. If the left operand is of a pointer type, the right operand must be of a pointer type, or it must be a constant expression that evaluates to 0. When the left operand is of an integral type, the right operand must not be of a pointer type.

## Result of assignment operators

The assignment operators return the value of the object specified by the left operand after the assignment. The resultant type is the type of the left operand. The result of an assignment expression is always an l-value. These operators have right-to-left associativity. The left operand must be a modifiable l-value.

In ANSI C, the result of an assignment expression isn't an l-value. That means the legal C++ expression `(a += b) += c` isn't allowed in C.