Given a hexadecimal string in Swift, convert to hex value - swift

Question

Suppose I am given a string like this:
D7C17A4F
How do I convert each individual character to a hex value?
So D should be 0xD, 7 should be 0x7…
Right now, I have each individual character represented as it's ASCII value. D is 68, 7 is 55. I'm trying to pack those two values into one byte. For example: D7 becomes 0xD7 and C1 becomes 0xC1. I can't do that using the ASCII decimal values though.

Answer 1

A possible solution:
let string = "D7C17A4F"
let chars = Array(string)
let numbers = map (stride(from: 0, to: chars.count, by: 2)) {
strtoul(String(chars[$0 ..< $0+2]), nil, 16)
}
Using the approach from https://stackoverflow.com/a/29306523/1187415,
the string is split into substrings of two characters.
Each substring is interpreted as a sequence of digits
in base 16, and converted to a number with strtoul().
Verify the result:
println(numbers)
// [215, 193, 122, 79]
println(map(numbers, { String(format: "%02X", $0) } ))
// [D7, C1, 7A, 4F]
Update for Swift 2 (Xcode 7):
let string = "D7C17A4F"
let chars = Array(string.characters)
let numbers = 0.stride(to: chars.count, by: 2).map {
UInt8(String(chars[$0 ..< $0+2]), radix: 16) ?? 0
}
print(numbers)
or
let string = "D7C17A4F"
var numbers = [UInt8]()
var from = string.startIndex
while from != string.endIndex {
let to = from.advancedBy(2, limit: string.endIndex)
numbers.append(UInt8(string[from ..< to], radix: 16) ?? 0)
from = to
}
print(numbers)
The second solution looks a bit more complicated but has the small
advantage that no additional chars array is needed.

Answer 2

Swift 3 version, modified from #Martin R's answer. This variant also accepts incoming string with odd length.
let string = "D7C17A4F"
let chars = Array(string.characters)
let numbers = stride(from: 0, to: chars.count, by: 2).map() {
strtoul(String(chars[$0 ..< min($0 + 2, chars.count)]), nil, 16)
}

Answer 3

Use chunks!
"D7C17A4F"
.chunks(ofCount: 2)
.map { UInt8($0, radix: 0x10)! }

Answer 4

My variation of #martin-r answer:
extension String {
func hexToByteArray() -> [UInt8] {
let byteCount = self.utf8.count / 2
var array = [UInt8](count: byteCount, repeatedValue: 0)
var from = self.startIndex
for i in 0..<byteCount {
let to = from.successor()
let sub = self.substringWithRange(from...to)
array[i] = UInt8(sub, radix: 16) ?? 0
from = to.successor()
}
return array
}
}

Answer 5

here is the more generic, "pure swift" approach (no Foundation required :-))
extension UnsignedInteger {
var hex: String {
var str = String(self, radix: 16, uppercase: true)
while str.characters.count < 2 * MemoryLayout<Self>.size {
str.insert("0", at: str.startIndex)
}
return str
}
}
extension Array where Element: UnsignedInteger {
var hex: String {
var str = ""
self.forEach { (u) in
str.append(u.hex)
}
return str
}
}
let str = [UInt8(1),22,63,41].hex // "01163F29"
let str2 = [UInt(1),22,63,41].hex // "00000000000000010000000000000016000000000000003F0000000000000029"
extension String {
func toUnsignedInteger<T:UnsignedInteger>()->[T]? {
var ret = [T]()
let nibles = MemoryLayout<T>.size * 2
for i in stride(from: 0, to: characters.count, by: nibles) {
let start = self.index(startIndex, offsetBy: i)
guard let end = self.index(start, offsetBy: nibles, limitedBy: endIndex),
let ui = UIntMax(self[start..<end], radix: 16) else { return nil }
ret.append(T(ui))
}
return ret
}
}
let u0:[UInt8]? = str.toUnsignedInteger() // [1, 22, 63, 41]
let u1 = "F2345f".toUnsignedInteger() as [UInt8]? // [18, 52, 95]
let u2 = "12345f".toUnsignedInteger() as [UInt16]? // nil
let u3 = "12345g".toUnsignedInteger() as [UInt8]? // nil
let u4 = "12345f".toUnsignedInteger() as [UInt]? // nil
let u5 = "12345678".toUnsignedInteger() as [UInt8]? // [18, 52, 86, 120]
let u6 = "12345678".toUnsignedInteger() as [UInt16]? // [4660, 22136]
let u7 = "1234567812345678".toUnsignedInteger() as [UInt]? // [1311768465173141112]
It is very easily to do the same for SignedInteger as well, but better approach will be to map results to signed type
let u8 = u1?.map { Int8(bitPattern: $0) } // [-14, 52, 95]