Assignment operators

Assignment and compound assignment operators are binary operators that modify the variable to their left using the value to their right.

Operator Operator name Example Description Equivalent of
= basic assignment a = b a becomes equal to b N/A
+ = addition assignment a + = b a becomes equal to the addition of a and b a = a + b
- = subtraction assignment a - = b a becomes equal to the subtraction of b from a a = a - b
* = multiplication assignment a * = b a becomes equal to the product of a and b a = a * b
/ = division assignment a / = b a becomes equal to the division of a by b a = a / b
% = modulo assignment a % = b a becomes equal to the remainder of a divided by b a = a % b
& = bitwise AND assignment a & = b a becomes equal to the bitwise AND of a and b a = a & b
| = bitwise OR assignment a | = b a becomes equal to the bitwise OR of a and b a = a | b
^ = bitwise XOR assignment a ^ = b a becomes equal to the bitwise XOR of a and b a = a ^ b
bitwise left shift assignment a a becomes equal to a left shifted by b a = a
>>= bitwise right shift assignment a >>= b a becomes equal to a right shifted by b a = a >> b

Contents

[edit] Simple assignment

The simple assignment operator expressions have the form

lhs = rhs
lhs - modifiable lvalue expression of any complete object type
rhs - expression of any type implicitly convertible to lhs or compatible with lhs

Assignment performs implicit conversion from the value of rhs to the type of lhs and then replaces the value in the object designated by lhs with the converted value of rhs .

Assignment also returns the same value as what was stored in lhs (so that expressions such as a = b = c are possible). The value category of the assignment operator is non-lvalue (so that expressions such as ( a = b ) = c are invalid).

rhs and lhs must satisfy one of the following:

[edit] Notes

If rhs and lhs overlap in memory (e.g. they are members of the same union), the behavior is undefined unless the overlap is exact and the types are compatible.

Although arrays are not assignable, an array wrapped in a struct is assignable to another object of the same (or compatible) struct type.

The side effect of updating lhs is sequenced after the value computations, but not the side effects of lhs and rhs themselves and the evaluations of the operands are, as usual, unsequenced relative to each other (so the expressions such as i = ++ i ; are undefined)

Assignment strips extra range and precision from floating-point expressions (see FLT_EVAL_METHOD ).

In C++, assignment operators are lvalue expressions, not so in C.

Run this code
#include int main(void) { // integers int i = 1, j = 2, k = 3; // initialization, not assignment i = j = k; // values of i and j are now 3 // (i = j) = k; // Error: lvalue required printf("%d %d %d\n", i, j, k); // pointers const char c = 'A'; // initialization; not assignment const char *p = &c; // initialization; not assignment const char **cpp = &p; // initialization; not assignment // cpp = &p; // Error: char** is not convertible to const char** *cpp = &c; // OK, char* is convertible to const char* printf("%c \n", **cpp); cpp = 0; // OK, null pointer constant is convertible to any pointer // arrays int arr1[2] = {1,2}, arr2[2] = {3, 4}; // arr1 = arr2; // Error: cannot assign to an array printf("arr1[0]=%d arr1[1]=%d arr2[0]=%d arr2[1]=%d\n", arr1[0], arr1[1], arr2[0], arr2[1]); struct { int arr[2]; } sam1 = { {5, 6} }, sam2 = { {7, 8} }; sam1 = sam2; // OK: can assign arrays wrapped in structs printf("%d %d \n", sam1.arr[0], sam1.arr[1]); }
3 3 3 A arr1[0]=1 arr1[1]=2 arr2[0]=3 arr2[1]=4 7 8

[edit] Compound assignment

The compound assignment operator expressions have the form

lhs op rhs
op - one of * = , / = % = , + = - = , >= , & = , ^ = , | =
lhs , rhs - expressions with arithmetic types (where lhs may be qualified or atomic), except when op is + = or - = , which also accept pointer types with the same restrictions as + and -

The expression lhs @= rhs is exactly the same as lhs = lhs @ ( rhs ) , except that lhs is evaluated only once.

If lhs has atomic type, the operation behaves as a single atomic read-modify-write operation with memory order memory_order_seq_cst .

For integer atomic types, the compound assignment @ = is equivalent to:

T1* addr = &lhs; T2 val = rhs; T1 old = *addr; T1 new; do { new = old @ val } while (!atomic_compare_exchange_strong(addr, &old, new);
Run this code
#include int main(void) { int x = 10; int hundred = 100; int ten = 10; int fifty = 50; printf("%d %d %d %d\n", x, hundred, ten, fifty); hundred *= x; ten /= x; fifty %= x; printf("%d %d %d %d\n", x, hundred, ten, fifty); return 0; }
10 100 10 50 10 1000 1 0

[edit] References

[edit] See Also

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

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

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

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

_Alignof
(since C11)