Advanced Operators
In addition to the operators described in Basic Operators, Swift provides several advanced operators that perform more complex value manipulation. These include all of the bitwise and bit shifting operators you will be familiar with from C and ObjectiveC.
Unlike arithmetic operators in C, arithmetic operators in Swift do not overflow by default. Overflow behavior is trapped and reported as an error. To opt in to overflow behavior, use Swift’s second set of arithmetic operators that overflow by default, such as the overflow addition operator (&+
). All of these overflow operators begin with an ampersand (&
).
When you define your own structures, classes, and enumerations, it can be useful to provide your own implementations of the standard Swift operators for these custom types. Swift makes it easy to provide tailored implementations of these operators and to determine exactly what their behavior should be for each type you create.
You’re not limited to the predefined operators. Swift gives you the freedom to define your own custom infix, prefix, postfix, and assignment operators, with custom precedence and associativity values. These operators can be used and adopted in your code like any of the predefined operators, and you can even extend existing types to support the custom operators you define.
Bitwise Operators
Bitwise operators enable you to manipulate the individual raw data bits within a data structure. They are often used in lowlevel programming, such as graphics programming and device driver creation. Bitwise operators can also be useful when you work with raw data from external sources, such as encoding and decoding data for communication over a custom protocol.
Swift supports all of the bitwise operators found in C, as described below.
Bitwise NOT Operator
The bitwise NOT operator (~
) inverts all bits in a number:
The bitwise NOT operator is a prefix operator, and appears immediately before the value it operates on, without any white space:
let initialBits: UInt8 = 0b00001111
let invertedBits = ~initialBits // equals 11110000
UInt8 integers have eight bits and can store any value between 0
and 255
. This example initializes a UInt8
integer with the binary value 00001111
, which has its first four bits set to 0
, and its second four bits set to 1
. This is equivalent to a decimal value of 15
.
The bitwise NOT operator is then used to create a new constant called invertedBits
, which is equal to initialBits
, but with all of the bits inverted. Zeros become ones, and ones become zeros. The value of invertedBits
is 11110000
, which is equal to an unsigned decimal value of 240
.
Bitwise AND Operator
The bitwise AND operator (&
) combines the bits of two numbers. It returns a new number whose bits are set to 1
only if the bits were equal to 1
in both input numbers:
In the example below, the values of firstSixBits
and lastSixBits
both have four middle bits equal to 1
. The bitwise AND operator combines them to make the number 00111100
, which is equal to an unsigned decimal value of 60
:
let firstSixBits: UInt8 = 0b11111100
let lastSixBits: UInt8 = 0b00111111
let middleFourBits = firstSixBits & lastSixBits // equals 00111100
Bitwise OR Operator
The bitwise OR operator (
) compares the bits of two numbers. The operator returns a new number whose bits are set to 1
if the bits are equal to 1
in either input number:
In the example below, the values of someBits
and moreBits
have different bits set to 1
. The bitwise OR operator combines them to make the number 11111110
, which equals an unsigned decimal of 254
:
let someBits: UInt8 = 0b10110010
let moreBits: UInt8 = 0b01011110
let combinedbits = someBits  moreBits // equals 11111110
Bitwise XOR Operator
The bitwise XOR operator, or “exclusive OR operator” (^
), compares the bits of two numbers. The operator returns a new number whose bits are set to 1
where the input bits are different and are set to 0
where the input bits are the same:
In the example below, the values of firstBits
and otherBits
each have a bit set to 1
in a location that the other does not. The bitwise XOR operator sets both of these bits to 1
in its output value. All of the other bits in firstBits
and otherBits
match and are set to 0
in the output value:
let firstBits: UInt8 = 0b00010100
let otherBits: UInt8 = 0b00000101
let outputBits = firstBits ^ otherBits // equals 00010001
Bitwise Left and Right Shift Operators
The bitwise left shift operator (<<
) and bitwise right shift operator (>>
) move all bits in a number to the left or the right by a certain number of places, according to the rules defined below.
Bitwise left and right shifts have the effect of multiplying or dividing an integer by a factor of two. Shifting an integer’s bits to the left by one position doubles its value, whereas shifting it to the right by one position halves its value.
Shifting Behavior for Unsigned Integers
The bitshifting behavior for unsigned integers is as follows:
 Existing bits are moved to the left or right by the requested number of places.
 Any bits that are moved beyond the bounds of the integer’s storage are discarded.
 Zeros are inserted in the spaces left behind after the original bits are moved to the left or right.
This approach is known as a logical shift.
The illustration below shows the results of 11111111 << 1
(which is 11111111
shifted to the left by 1
place), and 11111111 >> 1
(which is 11111111
shifted to the right by 1
place). Blue numbers are shifted, gray numbers are discarded, and orange zeros are inserted:
Here’s how bit shifting looks in Swift code:
let shiftBits: UInt8 = 4 // 00000100 in binary
shiftBits << 1 // 00001000
shiftBits << 2 // 00010000
shiftBits << 5 // 10000000
shiftBits << 6 // 00000000
shiftBits >> 2 // 00000001
You can use bit shifting to encode and decode values within other data types:
let pink: UInt32 = 0xCC6699
let redComponent = (pink & 0xFF0000) >> 16 // redComponent is 0xCC, or 204
let greenComponent = (pink & 0x00FF00) >> 8 // greenComponent is 0x66, or 102
let blueComponent = pink & 0x0000FF // blueComponent is 0x99, or 153
This example uses a UInt32
constant called pink to store a Cascading Style Sheets color value for the color pink. The CSS color value #CC6699
is written as 0xCC6699
in Swift’s hexadecimal number representation. This color is then decomposed into its red (CC
), green (66
), and blue (99
) components by the bitwise AND operator (&
) and the bitwise right shift operator (>>
).
The red component is obtained by performing a bitwise AND between the numbers 0xCC6699
and 0xFF0000
. The zeros in 0xFF0000
effectively “mask” the second and third bytes of 0xCC6699
, causing the 6699
to be ignored and leaving 0xCC0000
as the result.
This number is then shifted 16 places to the right (>> 16
). Each pair of characters in a hexadecimal number uses 8 bits, so a move 16 places to the right will convert 0xCC0000
into 0x0000CC
. This is the same as 0xCC
, which has a decimal value of 204
.
Similarly, the green component is obtained by performing a bitwise AND between the numbers 0xCC6699
and 0x00FF00
, which gives an output value of 0x006600
. This output value is then shifted eight places to the right, giving a value of 0x66
, which has a decimal value of 102
.
Finally, the blue component is obtained by performing a bitwise AND between the numbers 0xCC6699
and 0x0000FF
, which gives an output value of 0x000099
. There’s no need to shift this to the right, as 0x000099
already equals 0x99
, which has a decimal value of 153
.
Shifting Behavior for Signed Integers
The shifting behavior is more complex for signed integers than for unsigned integers, because of the way signed integers are represented in binary. (The examples below are based on 8bit signed integers for simplicity, but the same principles apply for signed integers of any size.)
Signed integers use their first bit (known as the sign bit) to indicate whether the integer is positive or negative. A sign bit of 0
means positive, and a sign bit of 1
means negative.
The remaining bits (known as the value bits) store the actual value. Positive numbers are stored in exactly the same way as for unsigned integers, counting upwards from 0
. Here’s how the bits inside an Int8
look for the number 4
:
The sign bit is 0
(meaning “positive”), and the seven value bits are just the number 4
, written in binary notation.
Negative numbers, however, are stored differently. They are stored by subtracting their absolute value from 2
to the power of n
, where n is the number of value bits. An eightbit number has seven value bits, so this means 2
to the power of 7
, or 128
.
Here’s how the bits inside an Int8
look for the number 4
:
This time, the sign bit is 1
(meaning “negative”), and the seven value bits have a binary value of 124
(which is 128  4
):
This encoding for negative numbers is known as a two’s complement representation. It may seem an unusual way to represent negative numbers, but it has several advantages.
First, you can add 1
to 4
, simply by performing a standard binary addition of all eight bits (including the sign bit), and discarding anything that doesn’t fit in the eight bits once you’re done:
Second, the two’s complement representation also lets you shift the bits of negative numbers to the left and right like positive numbers, and still end up doubling them for every shift you make to the left, or halving them for every shift you make to the right. To achieve this, an extra rule is used when signed integers are shifted to the right: When you shift signed integers to the right, apply the same rules as for unsigned integers, but fill any empty bits on the left with the sign bit, rather than with a zero.
This action ensures that signed integers have the same sign after they are shifted to the right, and is known as an arithmetic shift.
Because of the special way that positive and negative numbers are stored, shifting either of them to the right moves them closer to zero. Keeping the sign bit the same during this shift means that negative integers remain negative as their value moves closer to zero.
Overflow Operators
If you try to insert a number into an integer constant or variable that cannot hold that value, by default Swift reports an error rather than allowing an invalid value to be created. This behavior gives extra safety when you work with numbers that are too large or too small.
For example, the Int16
integer type can hold any signed integer between 32768
and 32767
. Trying to set an Int16
constant or variable to a number outside of this range causes an error:
var potentialOverflow = Int16.max
// potentialOverflow equals 32767, which is the maximum value an Int16 can hold
potentialOverflow += 1
// this causes an error
Providing error handling when values get too large or too small gives you much more flexibility when coding for boundary value conditions.
However, when you specifically want an overflow condition to truncate the number of available bits, you can opt in to this behavior rather than triggering an error. Swift provides three arithmetic overflow operators that opt in to the overflow behavior for integer calculations. These operators all begin with an ampersand (&
):

Overflow addition (
&+
) 
Overflow subtraction (
&
) 
Overflow multiplication (
&*
)
Value Overflow
Numbers can overflow in both the positive and negative direction.
Here’s an example of what happens when an unsigned integer is allowed to overflow in the positive direction, using the overflow addition operator (&+
):
var unsignedOverflow = UInt8.max
// unsignedOverflow equals 255, which is the maximum value a UInt8 can hold
unsignedOverflow = unsignedOverflow &+ 1
// unsignedOverflow is now equal to 0
The variable unsignedOverflow
is initialized with the maximum value a UInt8
can hold (255
, or 11111111
in binary). It is then incremented by 1
using the overflow addition operator (&+
). This pushes its binary representation just over the size that a UInt8
can hold, causing it to overflow beyond its bounds, as shown in the diagram below. The value that remains within the bounds of the UInt8
after the overflow addition is 00000000
, or zero.
Something similar happens when an unsigned integer is allowed to overflow in the negative direction. Here’s an example using the overflow subtraction operator (&
):
var unsignedOverflow = UInt8.min
// unsignedOverflow equals 0, which is the minimum value a UInt8 can hold
unsignedOverflow = unsignedOverflow & 1
// unsignedOverflow is now equal to 255
The minimum value that a UInt8
can hold is zero, or 00000000
in binary. If you subtract 1
from 00000000
using the overflow subtraction operator (&
), the number will overflow and wrap around to 11111111
, or 255
in decimal.
Overflow also occurs for signed integers. All addition and subtraction for signed integers is performed in bitwise fashion, with the sign bit included as part of the numbers being added or subtracted, as described in Bitwise Left and Right Shift Operators.
var signedOverflow = Int8.min
// signedOverflow equals 128, which is the minimum value an Int8 can hold
signedOverflow = signedOverflow & 1
// signedOverflow is now equal to 127
The minimum value that an Int8
can hold is 128
, or 10000000
in binary. Subtracting 1
from this binary number with the overflow operator gives a binary value of 01111111
, which toggles the sign bit and gives positive 127
, the maximum positive value that an Int8
can hold.
For both signed and unsigned integers, overflow in the positive direction wraps around from the maximum valid integer value back to the minimum, and overflow in the negative direction wraps around from the minimum value to the maximum.
Precedence and Associativity
Operator precedence gives some operators higher priority than others; these operators are applied first.
Operator associativity defines how operators of the same precedence are grouped together — either grouped from the left, or grouped from the right. Think of it as meaning “they associate with the expression to their left,” or “they associate with the expression to their right.”
It is important to consider each operator’s precedence and associativity when working out the order in which a compound expression will be calculated. For example, operator precedence explains why the following expression equals 17
.
2 + 3 % 4 * 5
// this equals 17
If you read strictly from left to right, you might expect the expression to be calculated as follows:
 2 plus 3 equals 5
 5 remainder 4 equals 1
 1 times 5 equals 5
However, the actual answer is 17
, not 5
. Higherprecedence operators are evaluated before lowerprecedence ones. In Swift, as in C, the remainder operator (%
) and the multiplication operator (*
) have a higher precedence than the addition operator (+
). As a result, they are both evaluated before the addition is considered.
However, remainder and multiplication have the same precedence as each other. To work out the exact evaluation order to use, you also need to consider their associativity. Remainder and multiplication both associate with the expression to their left. Think of this as adding implicit parentheses around these parts of the expression, starting from their left:
2 + ((3 % 4) * 5)
3 % 4
) is 3
, so this is equivalent to:2 + 3 * 5
(3 * 5)
is 15
, so this is equivalent to:2 + 15
This calculation yields the final answer of 17
.
For information about the operators provided by the Swift standard library, including a complete list of the operator precedence groups and associativity settings, see Operator Declarations.
Swift’s operator precedences and associativity rules are simpler and more predictable than those found in C and ObjectiveC. However, this means that they are not exactly the same as in Cbased languages. Be careful to ensure that operator interactions still behave in the way you intend when porting existing code to Swift.