What's the best way to cut Swift string into 2-letter-strings?

Question

I need to split a string into 2-letter pieces. Like “friend\" -> \"fr\" \"ie\" \"nd\". (Okay, its a step for me to change HEX string to Uint8 Array)

My code is

Answer 1

Another option just for fun:

extension String {
    var pairs:[String] {
        var result:[String] = []
        let chars = Array(characters)
        for index in 0.stride(to: chars.count, by: 2) {
            result.append(String(chars[index..<min(index+2, chars.count)]))
        }
        return result
    }
}
let input = "friends"
let pairs = input.pairs
print(pairs) // ["fr", "ie", "nd", "s"]

Answer 2

Your range wasn't working because you need to use ... instead of ..<.

let input = "ff103"
var bytes = [UInt8]()

let strlen = input.characters.count
for i in 0 ..< (strlen + 1)/2 {
    let str = input[input.startIndex.advancedBy(i*2)...input.startIndex.advancedBy(min(strlen - 1, i*2+1))]
    bytes.append(UInt8(str,radix: 16) ?? 0)
}

print(bytes)  // [255, 16, 3]

---

Here is another take on splitting the string into 2-letter strings. advancedBy() is an expensive O(n) operation, so this version keeps track of start and just marches it ahead by 2 each loop, and end is based on start:

let input = "friends"
var strings = [String]()

let strlen = input.characters.count
var start = input.startIndex
let lastIndex = strlen > 0 ? input.endIndex.predecessor() : input.startIndex

for i in 0 ..< (strlen + 1)/2 {
    start = i > 0 ? start.advancedBy(2) : start
    let end = start < lastIndex ? start.successor() : start
    let str = input[start...end]
    strings.append(str)
}

print(strings) // ["fr", "ie", "nd", "s"]

---

Alternate Answer:

Using ranges is probably overkill. It is easy just to add the characters to an array and make Strings from those:

let input = "friends"
var strings = [String]()
var newchars = [Character]()

for c in input.characters {
    newchars.append(c)
    if newchars.count == 2 {
        strings.append(String(newchars))
        newchars = []
    }
}

if newchars.count > 0 {
    strings.append(String(newchars))
}

print(strings) // ["fr", "ie", "nd", "s"]

---

And here is the new version for making [UInt8]:

let input = "ff103"
var bytes = [UInt8]()
var newchars = [Character]()

for c in input.characters {
    newchars.append(c)
    if newchars.count == 2 {
        bytes.append(UInt8(String(newchars), radix: 16) ?? 0)
        newchars = []
    }
}

if newchars.count > 0 {
    bytes.append(UInt8(String(newchars), radix: 16) ?? 0)
}

print(bytes) // [255, 16, 3]

---

Based on @LeoDabus' answer, we can make an extension with a method that will return substrings of any length, and a computed property that returns [UInt8]:

extension String {
    func substringsOfLength(length: Int) -> [String] {
        if length < 1 { return [] }

        var result:[String] = []
        let chars = Array(characters)
        for index in 0.stride(to: chars.count, by: length) {
            result.append(String(chars[index ..< min(index+length, chars.count)]))
        }
        return result
    }

    var toUInt8: [UInt8] {
        var result:[UInt8] = []
        let chars = Array(characters)
        for index in 0.stride(to: chars.count, by: 2) {
            let str = String(chars[index ..< min(index+2, chars.count)])
            result.append(UInt8(str, radix: 16) ?? 0)
        }
        return result
    }
}

let input = "friends"
let str2 = input.substringsOfLength(2)  // ["fr", "ie", "nd", "s"]
let str0 = input.substringsOfLength(0)  // []
let str3 = input.substringsOfLength(3)  // ["fri", "end", "s"]

let bytes = "ff107".toUInt8  // [255, 16, 7]