How to convert an Int to Hex String in Swift
You can now do:
let n = 14
var st = String(format:"%02X", n)
st += " is the hexadecimal representation of \(n)"
print(st)
0E is the hexadecimal representation of 14
Note: The 2
in this example is the field width and represents the minimum length desired. The 0
tells it to pad the result with leading 0
's if necessary. (Without the 0
, the result would be padded with leading spaces). Of course, if the result is larger than two characters, the field length will not be clipped to a width of 2
; it will expand to whatever length is necessary to display the full result.
This only works if you have Foundation
imported (this includes the import of Cocoa
or UIKit
). This isn't a problem if you're doing iOS or macOS programming.
Use uppercase X
if you want A...F
and lowercase x
if you want a...f
:
String(format: "%x %X", 64206, 64206) // "face FACE"
If you want to print integer values larger than UInt32.max
, add ll
(el-el, not eleven) to the format string:
let n = UInt64.max
print(String(format: "%llX is hexadecimal for \(n)", n))
FFFFFFFFFFFFFFFF is hexadecimal for 18446744073709551615
Original Answer
You can still use NSString
to do this. The format is:
var st = NSString(format:"%2X", n)
This makes st
an NSString
, so then things like +=
do not work. If you want to be able to append to the string with +=
make st
into a String
like this:
var st = NSString(format:"%2X", n) as String
or
var st = String(NSString(format:"%2X", n))
or
var st: String = NSString(format:"%2X", n)
Then you can do:
let n = 123
var st = NSString(format:"%2X", n) as String
st += " is the hexadecimal representation of \(n)"
// "7B is the hexadecimal representation of 123"
With Swift 5, according to your needs, you may choose one of the three following methods in order to solve your problem.
#1. Using String
's init(_:radix:uppercase:)
initializer
Swift String
has a init(_:radix:uppercase:)
initializer with the following declaration:
init<T>(_ value: T, radix: Int = 10, uppercase: Bool = false) where T : BinaryInteger
Creates a string representing the given value in base 10, or some other specified base.
The Playground code below shows how to create a String
instance that represents an integer value in hexadecimal format by using init(_:radix:uppercase:)
and without having to import Foundation
:
let string1 = String(2, radix: 16)
print(string1) // prints: "2"
let string2 = String(211, radix: 16)
print(string2) // prints: "d3"
let string3 = String(211, radix: 16, uppercase: true)
print(string3) // prints: "D3"
#2. Using String
's init(format:_:)
initializer
Foundation
provides String
a init(format:_:)
initializer. init(format:_:)
has the following declaration:
init(format: String, _ arguments: CVarArg...)
Returns a
String
object initialized by using a given format string as a template into which the remaining argument values are substituted.
The Apple's String Programming Guide gives a list of the format specifiers that are supported by String
and NSString
. Among those format specifiers, %X
has the following description:
Unsigned 32-bit integer (
unsigned int
), printed in hexadecimal using the digits 0–9 and uppercase A–F.
The Playground code below shows how to create a String
instance that represents an integer value in hexadecimal format with init(format:_:)
:
import Foundation
let string1 = String(format:"%X", 2)
print(string1) // prints: "2"
let string2 = String(format:"%02X", 1)
print(string2) // prints: "01"
let string3 = String(format:"%02X", 211)
print(string3) // prints: "D3"
let string4 = String(format: "%02X, %02X, %02X", 12, 121, 255)
print(string4) // prints: "0C, 79, FF"
#3. Using String
's init(format:arguments:)
initializer
Foundation
provides String
a init(format:arguments:)
initializer. init(format:arguments:)
has the following declaration:
init(format: String, arguments: [CVarArg])
Returns a
String
object initialized by using a given format string as a template into which the remaining argument values are substituted according to the user’s default locale.
The Playground code below shows how to create a String
instance that represents an integer value in hexadecimal format with init(format:arguments:)
:
import Foundation
let string1 = String(format:"%X", arguments: [2])
print(string1) // prints: "2"
let string2 = String(format:"%02X", arguments: [1])
print(string2) // prints: "01"
let string3 = String(format:"%02X", arguments: [211])
print(string3) // prints: "D3"
let string4 = String(format: "%02X, %02X, %02X", arguments: [12, 121, 255])
print(string4) // prints: "0C, 79, FF"
Swift 5.2.4
let value = 200
let hexString = String(format: "%02X", value)
In Swift there is a specific init
method on String
for exactly this:
let hex = String(0xF, radix: 16, uppercase: false)
println("hex=\(hex)") // Output: f