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 Objective-C.
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 enable you to manipulate the individual raw data bits within a data structure. They are often used in low-level 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
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
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
11110000, which is equal to an unsigned decimal value of
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
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
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
moreBits have different bits set to
1. The bitwise OR operator combines them to make the number
11111110, which equals an unsigned decimal of
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
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
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 bit-shifting 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
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
0x0000CC. This is the same as
0xCC, which has a decimal value of
Similarly, the green component is obtained by performing a bitwise AND between the numbers
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
Finally, the blue component is obtained by performing a bitwise AND between the numbers
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
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 8-bit 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
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 eight-bit number has seven value bits, so this means
2 to the power of
Here’s how the bits inside an
Int8 look for the number
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
-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.
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
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 (
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
unsignedOverflow is initialized with the maximum value a
UInt8 can hold (
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
00000000 using the overflow subtraction operator (
&-), the number will overflow and wrap around to
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
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
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
5. Higher-precedence operators are evaluated before lower-precedence 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
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 Objective-C. However, this means that they are not exactly the same as in C-based languages. Be careful to ensure that operator interactions still behave in the way you intend when porting existing code to Swift.