In Obj-C I used to convert an unsigned integer n to a hex string with
NSString *st = [NSString stringWithFormat:@"%2X", n];
I tried for a long time to translate this into Swift language, but unsuccessfully.
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"