Related
I have a double:
let value = 0.99720317490866084
And a Double extention function:
extension Double {
func stringWithFixedFractionDigits(min: Int, max: Int) -> String {
let formatter = NumberFormatter()
formatter.minimumFractionDigits = min
formatter.maximumFractionDigits = max
formatter.minimumIntegerDigits = 1
let numberObject = NSNumber(value: self)
return formatter.string(from: numberObject) ?? "\(self)"
}
}
If I use:
value.stringWithFixedFractionDigits(min: 2, max: 2)
I get 1.00 but I would like to get 0.99
What can I change?
You just need to set your NumberFormatter rounding mode property to .down:
formatter.roundingMode = .down
Note that you don't need to create a new NSNumber object, you can safely cast from Double to NSNumber or use string(for: Any) method instead. As an alternative you can extend the protocol FloatingPoint to make it available to all Float types :
extension Formatter {
static let number = NumberFormatter()
}
extension FloatingPoint {
func formattedWithFractionDigits(minimum: Int = 2, maximum: Int = 2, minimumIntegerDigits: Int = 1, roundingMode: NumberFormatter.RoundingMode = .halfEven) -> String {
Formatter.number.roundingMode = roundingMode
Formatter.number.minimumFractionDigits = minimum
Formatter.number.maximumFractionDigits = maximum
Formatter.number.minimumIntegerDigits = minimumIntegerDigits
return Formatter.number.string(for: self) ?? ""
}
}
0.9972031749.formattedWithFractionDigits() // 1.00
0.9972031749.formattedWithFractionDigits(roundingMode: .down) // "0.99"
0.9972031749.formattedWithFractionDigits(minimumIntegerDigits: 0, roundingMode: .down) // ".99"
everyone ! Lets say we have
let random = arc4random_uniform(6)
how do i make it not repeat the same number more then two times ? I tried doing it like this :
let previousNumber = Int()
let lastNumber = Int ()
let random = Int(arc4random_uniform(6))
if random == previousNumber {
lastNumber = previousNumber
} else {
previousNumber = random
}
if random == lastNumber {
random = Int(arc4random_uniform(6))
}
But it didn't work. I am new to swift and i didn't find a topic about this on the new swift 3 code. Thank you !
First of all lets build a class to save the recent history of the selected values
class History {
private let size: Int
private var values = [Int]()
init(size:Int) {
self.size = size
}
func add(value: Int) {
values.insert(value, at: 0)
if values.count > size {
values.removeLast()
}
}
var repeatedValueOnFullHistory: Int? {
guard Set(values).count <= 1 else { return nil }
return values.first
}
}
Next let build a Randomizer
class Randomizer {
private var allValues = [Int]()
private var history: History
init?(maxValue: Int) {
guard maxValue > 0 else { return nil }
self.allValues = Array(0...maxValue)
self.history = History(size: maxValue + 1)
}
var next: Int {
let excludedValue = history.repeatedValueOnFullHistory
let allowedValues = allValues.filter { excludedValue != $0 }
let randomIndex = Int(arc4random_uniform(UInt32(allowedValues.count)))
let nextValue = allowedValues[randomIndex]
history.add(value: nextValue)
return nextValue
}
}
And finally let test it
if let r = Randomizer(maxValue: 6) {
r.next // 6
r.next // 2
r.next // 1
r.next // 4
r.next // 6
r.next // 4
r.next // 1
}
I am attempting to build an anagram checker for swift. This is my code. In case you don't know an anagram checker checks if two strings have the same characters in them but, order does not matter.
func checkForAnagram(#firstString: String, #secondString: String) -> Bool {
var firstStringArray: [Character] = []
var secondStringArray: [Character] = []
/* if case matters delete the next four lines
and make sure your variables are not constants */
var first = firstString
var second = secondString
first = first.lowercaseString
second = second.lowercaseString
for charactersOne in first {
firstStringArray += [charactersOne]
}
for charactersTwo in second {
secondStringArray += [charactersTwo]
}
if firstStringArray.count != secondStringArray.count {
return false
} else {
for elements in firstStringArray {
if secondStringArray.contains(elements){
return true
} else {
return false
}
}
}
}
var a = "Hello"
var b = "oellh"
var c = "World"
checkForAnagram(firstString: a, secondString: b)
I am getting an error message of.
'[Character]' does not have a member 'contains'
The accepted answer is compact and elegant, but very inefficient if compared to other solutions.
I'll now propose and discuss the implementation of a few variants of anagram checker. To measure performance, I'll use the different variants to find the anagrams of a given word out of an array of 50,000+ words.
// Variant 1: Sorting of Character
// Measured time: 30.46 s
func anagramCheck1(a: String, b: String) -> Bool {
return a.characters.sorted() == b.characters.sorted()
}
This is essentially the solution of the accepted answer, written in Swift 3 syntax. It's very slow because Swift's String, unlike NSString, is based on Character, which handles Unicode characters properly.
A more efficient solution exploits the NSCountedSet class, which allows us to represent a string as a set of characters, each with its own count. Two strings are anagrams if they map to the same NSCountedSet.
Note: checking string lengths as a precondition makes the implementation always more efficient.
// Variant 2: NSCountedSet of Character
// Measured time: 4.81 s
func anagramCheck2(a: String, b: String) -> Bool {
guard a.characters.count == b.characters.count else { return false }
let aSet = NSCountedSet()
let bSet = NSCountedSet()
for c in a.characters {
aSet.add(c)
}
for c in b.characters {
bSet.add(c)
}
return aSet == bSet
}
Better but not excellent. Here, one of the "culprits" is the use of the native Swift Character type (from Swift's String). Moving back to good old Objective-C types (NSString and unichar) makes things more efficient.
// Variant 3: NSCountedSet of unichar
// Measured time: 1.31 s
func anagramCheck3(a: String, b: String) -> Bool {
let aString = a as NSString
let bString = b as NSString
let length = aString.length
guard length == bString.length else { return false }
let aSet = NSCountedSet()
let bSet = NSCountedSet()
for i in 0..<length {
aSet.add(aString.character(at: i))
bSet.add(bString.character(at: i))
}
return aSet == bSet
}
Using NSCountedSet is fine, but before we compare two NSCountedSet objects, we fully populate them. A useful alternative is to fully populate the NSCountedSet for only one of the two strings, and then, while we populate the NSCountedSet for the other string, we fail early if the other string contains a character that is not found in the NSCountedSet of the first string.
// Variant 4: NSCountedSet of unichar and early exit
// Measured time: 1.07 s
func anagramCheck4(a: String, b: String) -> Bool {
let aString = a as NSString
let bString = b as NSString
let length = aString.length
guard length == bString.length else { return false }
let aSet = NSCountedSet()
let bSet = NSCountedSet()
for i in 0..<length {
aSet.add(aString.character(at: i))
}
for i in 0..<length {
let c = bString.character(at: i)
if bSet.count(for: c) >= aSet.count(for: c) {
return false
}
bSet.add(c)
}
return true
}
This is about the best timing we are going to get (with Swift). However, for completeness, let me discuss one more variant of this kind.
The next alternative exploits a Swift Dictionary of type [unichar: Int] to store the number of repetitions for each character instead of NSCountedSet. It's slightly slower than the previous two variants, but we can reuse it later to obtain a faster implementation.
// Variant 5: counting repetitions with [unichar:Int]
// Measured time: 1.36
func anagramCheck5(a: String, b: String) -> Bool {
let aString = a as NSString
let bString = b as NSString
let length = aString.length
guard length == bString.length else { return false }
var aDic = [unichar:Int]()
var bDic = [unichar:Int]()
for i in 0..<length {
let c = aString.character(at: i)
aDic[c] = (aDic[c] ?? 0) + 1
}
for i in 0..<length {
let c = bString.character(at: i)
let count = (bDic[c] ?? 0) + 1
if count > aDic[c] ?? 0 {
return false
}
bDic[c] = count
}
return true
}
Note that a vanilla Objective-C implementation using NSCountedSet, corresponding to Variant 3, is faster than all the previous versions by a rather large margin.
// Variant 6: Objective-C and NSCountedSet
// Measured time: 0.65 s
- (BOOL)anagramChecker:(NSString *)a with:(NSString *)b {
if (a.length != b.length) {
return NO;
}
NSCountedSet *aSet = [[NSCountedSet alloc] init];
NSCountedSet *bSet = [[NSCountedSet alloc] init];
for (int i = 0; i < a.length; i++) {
[aSet addObject:#([a characterAtIndex:i])];
[bSet addObject:#([b characterAtIndex:i])];
}
return [aSet isEqual:bSet];
}
Another way we can improve upon the previous attempts is to observe that, if we need to find the anagram of a given word, we might as well consider that word as fixed, and thus we could build the corresponding structure (NSCountedSet, Dictionary, ...) for that word only once.
// Finding all the anagrams of word in words
// Variant 7: counting repetitions with [unichar:Int]
// Measured time: 0.58 s
func anagrams(word: String, from words: [String]) -> [String] {
let anagrammedWord = word as NSString
let length = anagrammedWord.length
var aDic = [unichar:Int]()
for i in 0..<length {
let c = anagrammedWord.character(at: i)
aDic[c] = (aDic[c] ?? 0) + 1
}
let foundWords = words.filter {
let string = $0 as NSString
guard length == string.length else { return false }
var bDic = [unichar:Int]()
for i in 0..<length {
let c = string.character(at: i)
let count = (bDic[c] ?? 0) + 1
if count > aDic[c] ?? 0 {
return false
}
bDic[c] = count
}
return true
}
return foundWords
}
Now, in the previous variant we have counted with a [unichar:Int] Dictionary. This proves slightly more efficient than using an NSCountedSet of unichar, either with early exit (0.60 s) or without (0.87 s).
You should try
func checkForAnagram(firstString firstString: String, secondString: String) -> Bool {
return firstString.lowercaseString.characters.sort() == secondString.lowercaseString.characters.sort()
}
func checkAnagrams(str1: String, str2: String) -> Bool {
guard str1.count == str2.count else { return false }
var dictionary = Dictionary<Character, Int>()
for index in 0..<str1.count {
let value1 = str1[str1.index(str1.startIndex, offsetBy: index)]
let value2 = str2[str2.index(str2.startIndex, offsetBy: index)]
dictionary[value1] = (dictionary[value1] ?? 0) + 1
dictionary[value2] = (dictionary[value2] ?? 0) - 1
}
return !dictionary.contains(where: {(_, value) in
return value != 0
})
}
Time complexity - O(n)
// Make sure name your variables correctly so you won't confuse
// Mutate the constants parameter, lowercase to handle capital letters and the sorted them to compare both. Finally check is there are equal return true or false.
func anagram(str1: String, srt2: String)->Bool{
let string1 = str1.lowercased().sorted()
let string2 = srt2.lowercased().sorted()
if string1 == string2 {
return true
}
return false
}
// This answer also would work
// Convert your parameters on Array, then sorted them and compare them
func ana(str1: String, str2: String)->Bool{
let a = Array(str1)
let b = Array(str2)
if a.sorted() == b.sorted() {
return true
}
return false
}
Don't forget whitespaces
func isAnagram(_ stringOne: String, stringTwo: String) -> Bool {
return stringOne.lowercased().sorted().filter { $0 != " "} stringTwo.lowercased().sorted().filter { $0 != " "}
}
Swift 4.1 Function will give you 3 questions answer for Anagram :-
1. Input Strings (a,b) are Anagram ? //Bool
2. If not an Anagram then number of count require to change Characters in strings(a,b) to make them anagram ? // Int
3. If not an Anagram then list of Characters needs to be change in strings(a,b) to make them anagram ? // [Character]
STEP 1:- Copy and Paste below function in to your required class:-
//MARK:- Anagram checker
func anagramChecker(a:String,b:String) -> (Bool,Int,[Character]) {
var aCharacters = Array(a)
var bCharacters = Array(b)
var count = 0
var isAnagram = true
var replacementRequiredWords:[Character] = [Character]()
if aCharacters.count == bCharacters.count {
let listA = aCharacters.filter { !bCharacters.contains($0) }
for i in 0 ..< listA.count {
if !replacementRequiredWords.contains(listA[i]) {
count = count + 1
replacementRequiredWords.append(listA[i])
isAnagram = false
}
}
let listB = bCharacters.filter { !aCharacters.contains($0) }
for i in 0 ..< listB.count {
if !replacementRequiredWords.contains(listB[i]) {
count = count + 1
replacementRequiredWords.append(listB[i])
isAnagram = false
}
}
}else{
//cant be an anagram
count = -1
}
return (isAnagram,count,replacementRequiredWords)
}
STEP 2 :- Make two Input Strings for test
// Input Strings
var a = "aeb"
var b = "abs"
STEP 3:- Print results :-
print("isAnagram : \(isAnagram(a: a, b: b).0)")
print("number of count require to change strings in anagram : \(isAnagram(a: a, b: b).1)")//-1 will come in case of cant be a Anagram
print("list of Characters needs to be change : \(isAnagram(a: a, b: b).2)")
Results of above exercise:-
isAnagram : false
number of count require to change strings in anagram : 2
list of Characters needs to be change : ["e", "s"]
Hope this 10 minutes exercise will give some support to my Swift
family for solving Anagram related problems easily. :)
We can use dictionary to construct a new data structure container. Then compare the value by key/character of the string.
func anagram(str1: String, str2 : String) -> Bool {
var dict1 = [Character: Int]()
var dict2 = [Character:Int]()
for i in str1 {
if let count = dict1[i] {
dict1[i] = count + 1
} else {
dict1[i] = 1
}
}
for j in str2 {
if let count = dict2[j] {
dict2[j] = count + 1
} else {
dict2[j] = 1
}
}
return dict1 == dict2 ? true : false
}
// input -> "anna", "aann"
// The count will look like:
// ["a": 2, "n": 2] & ["a": 2, "n": 2]
// then return true
Another easy that I just realise doing an Anagram function in Swift 5.X
func checkForAnagram(firstString firstString: String, secondString: String) -> Bool {
return !firstString.isEmpty && firstString.sorted() == secondString.sorted()
}
class Solution {
func isAnagram(_ s: String, _ t: String) -> Bool {
guard s.count == t.count else { return false }
let dictS = s.reduce(into: [Character: Int]()) { $0[$1, default: 0] += 1 }
let dictT = t.reduce(into: [Character: Int]()) { $0[$1, default: 0] += 1 }
for letter in s {
if let count = dictS[letter] {
guard count == dictT[letter] else { return false }
}
}
return true
}
}
Check two strings are anagram using inout method in Swift
func checkAnagramString(str1: inout String, str2: inout String)-> Bool{
var result:Bool = false
str1 = str1.lowercased().trimmingCharacters(in: .whitespace)
str2 = str2.lowercased().trimmingCharacters(in: .whitespaces)
if (str1.count != str2.count) {
return result
}
for c in str1 {
if str2.contains(c){
result = true
}
else{
result = false
return result
}
}
return result
}
Call function to check strings are anagram or not
var str1 = "tommarvoloriddle"
var str2 = "iamlordvoldemort"
print(checkAnagramString(str1: &str1, str2: &str2)) //Output = true.
func isAnagram(word1: String, word2: String) -> Bool {
let set1 = Set(word1)
let set2 = Set(word2)
return set1 == set2
}
or
func isAnagram(word1: String,word2: String) -> Bool {
return word1.lowercased().sorted() == word2.lowercased().sorted()
}
I'm displaying a distance with one decimal, and I would like to remove this decimal in case it is equal to 0 (ex: 1200.0Km), how could I do that in swift?
I'm displaying this number like this:
let distanceFloat: Float = (currentUser.distance! as NSString).floatValue
distanceLabel.text = String(format: "%.1f", distanceFloat) + "Km"
Swift 3/4:
var distanceFloat1: Float = 5.0
var distanceFloat2: Float = 5.540
var distanceFloat3: Float = 5.03
extension Float {
var clean: String {
return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(self)
}
}
print("Value \(distanceFloat1.clean)") // 5
print("Value \(distanceFloat2.clean)") // 5.54
print("Value \(distanceFloat3.clean)") // 5.03
Swift 2 (Original answer)
let distanceFloat: Float = (currentUser.distance! as NSString).floatValue
distanceLabel.text = String(format: distanceFloat == floor(distanceFloat) ? “%.0f" : "%.1f", distanceFloat) + "Km"
Or as an extension:
extension Float {
var clean: String {
return self % 1 == 0 ? String(format: "%.0f", self) : String(self)
}
}
Use NSNumberFormatter:
let formatter = NumberFormatter()
formatter.minimumFractionDigits = 0
formatter.maximumFractionDigits = 2
// Avoid not getting a zero on numbers lower than 1
// Eg: .5, .67, etc...
formatter.numberStyle = .decimal
let nums = [3.0, 5.1, 7.21, 9.311, 600.0, 0.5677, 0.6988]
for num in nums {
print(formatter.string(from: num as NSNumber) ?? "n/a")
}
Returns:
3
5.1
7.21
9.31
600
0.57
0.7
extension is the powerful way to do it.
Extension:
Code for Swift 2 (not Swift 3 or newer):
extension Float {
var cleanValue: String {
return self % 1 == 0 ? String(format: "%.0f", self) : String(self)
}
}
Usage:
var sampleValue: Float = 3.234
print(sampleValue.cleanValue)
3.234
sampleValue = 3.0
print(sampleValue.cleanValue)
3
sampleValue = 3
print(sampleValue.cleanValue)
3
Sample Playground file is here.
Update of accepted answer for swift 3:
extension Float {
var cleanValue: String {
return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(self)
}
}
usage would just be:
let someValue: Float = 3.0
print(someValue.cleanValue) //prints 3
To format it to String, follow this pattern
let aFloat: Float = 1.123
let aString: String = String(format: "%.0f", aFloat) // "1"
let aString: String = String(format: "%.1f", aFloat) // "1.1"
let aString: String = String(format: "%.2f", aFloat) // "1.12"
let aString: String = String(format: "%.3f", aFloat) // "1.123"
To cast it to Int, follow this pattern
let aInt: Int = Int(aFloat) // "1"
When you use String(format: initializer, Swift will automatically round the final digit as needed based on the following number.
You can use an extension as already mentioned, this solution is a little shorter though:
extension Float {
var shortValue: String {
return String(format: "%g", self)
}
}
Example usage:
var sample: Float = 3.234
print(sample.shortValue)
Swift 5
for Double it's same as #Frankie's answer for float
var dec: Double = 1.0
dec.clean // 1
for the extension
extension Double {
var clean: String {
return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(self)
}
}
Swift 5.5 makes it easy
Just use the new formatted() api with a default FloatingPointFormatStyle:
let values: [Double] = [1.0, 4.5, 100.0, 7]
for value in values {
print(value.formatted(FloatingPointFormatStyle()))
}
// prints "1, 4.5, 100, 7"
In Swift 4 try this.
extension CGFloat{
var cleanValue: String{
//return String(format: 1 == floor(self) ? "%.0f" : "%.2f", self)
return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(format: "%.2f", self)//
}
}
//How to use - if you enter more then two-character after (.)point, it's automatically cropping the last character and only display two characters after the point.
let strValue = "32.12"
print(\(CGFloat(strValue).cleanValue)
Formatting with maximum fraction digits, without trailing zeros
This scenario is good when a custom output precision is desired.
This solution seems roughly as fast as NumberFormatter + NSNumber solution from MirekE, but one benefit could be that we're avoiding NSObject here.
extension Double {
func string(maximumFractionDigits: Int = 2) -> String {
let s = String(format: "%.\(maximumFractionDigits)f", self)
var offset = -maximumFractionDigits - 1
for i in stride(from: 0, to: -maximumFractionDigits, by: -1) {
if s[s.index(s.endIndex, offsetBy: i - 1)] != "0" {
offset = i
break
}
}
return String(s[..<s.index(s.endIndex, offsetBy: offset)])
}
}
(works also with extension Float, but not the macOS-only type Float80)
Usage: myNumericValue.string(maximumFractionDigits: 2) or myNumericValue.string()
Output for maximumFractionDigits: 2:
1.0 → "1"
0.12 → "0.12"
0.012 → "0.01"
0.0012 → "0"
0.00012 → "0"
Simple :
Int(floor(myFloatValue))
NSNumberFormatter is your friend
let distanceFloat: Float = (currentUser.distance! as NSString).floatValue
let numberFormatter = NSNumberFormatter()
numberFormatter.positiveFormat = "###0.##"
let distance = numberFormatter.stringFromNumber(NSNumber(float: distanceFloat))!
distanceLabel.text = distance + " Km"
Here's the full code.
let numberA: Float = 123.456
let numberB: Float = 789.000
func displayNumber(number: Float) {
if number - Float(Int(number)) == 0 {
println("\(Int(number))")
} else {
println("\(number)")
}
}
displayNumber(numberA) // console output: 123.456
displayNumber(numberB) // console output: 789
Here's the most important line in-depth.
func displayNumber(number: Float) {
Strips the float's decimal digits with Int(number).
Returns the stripped number back to float to do an operation with Float(Int(number)).
Gets the decimal-digit value with number - Float(Int(number))
Checks the decimal-digit value is empty with if number - Float(Int(number)) == 0
The contents within the if and else statements doesn't need explaining.
This might be helpful too.
extension Float {
func cleanValue() -> String {
let intValue = Int(self)
if self == 0 {return "0"}
if self / Float (intValue) == 1 { return "\(intValue)" }
return "\(self)"
}
}
Usage:
let number:Float = 45.23230000
number.cleanValue()
Maybe stringByReplacingOccurrencesOfString could help you :)
let aFloat: Float = 1.000
let aString: String = String(format: "%.1f", aFloat) // "1.0"
let wantedString: String = aString.stringByReplacingOccurrencesOfString(".0", withString: "") // "1"
I got the code to convert String to HEX-String in Objective-C:
- (NSString *) CreateDataWithHexString:(NSString*)inputString {
NSUInteger inLength = [inputString length];
unichar *inCharacters = alloca(sizeof(unichar) * inLength);
[inputString getCharacters:inCharacters range:NSMakeRange(0, inLength)];
UInt8 *outBytes = malloc(sizeof(UInt8) * ((inLength / 2) + 1));
NSInteger i, o = 0;
UInt8 outByte = 0;
for (i = 0; i < inLength; i++) {
UInt8 c = inCharacters[i];
SInt8 value = -1;
if (c >= '0' && c <= '9') value = (c - '0');
else if (c >= 'A' && c <= 'F') value = 10 + (c - 'A');
else if (c >= 'a' && c <= 'f') value = 10 + (c - 'a');
if (value >= 0) {
if (i % 2 == 1) {
outBytes[o++] = (outByte << 4) | value;
outByte = 0;
} else {
outByte = value;
}
} else {
if (o != 0) break;
}
}
NSData *a = [[NSData alloc] initWithBytesNoCopy:outBytes length:o freeWhenDone:YES];
NSString* newStr = [NSString stringWithUTF8String:[a bytes]];
return newStr;
}
I want the same in Swift. Can anybody translate this code in Swift, or is there any easy way to do this in Swift?
This is my hex string to Data routine:
extension String {
/// Create `Data` from hexadecimal string representation
///
/// This creates a `Data` object from hex string. Note, if the string has any spaces or non-hex characters (e.g. starts with '<' and with a '>'), those are ignored and only hex characters are processed.
///
/// - returns: Data represented by this hexadecimal string.
var hexadecimal: Data? {
var data = Data(capacity: count / 2)
let regex = try! NSRegularExpression(pattern: "[0-9a-f]{1,2}", options: .caseInsensitive)
regex.enumerateMatches(in: self, range: NSRange(startIndex..., in: self)) { match, _, _ in
let byteString = (self as NSString).substring(with: match!.range)
let num = UInt8(byteString, radix: 16)!
data.append(num)
}
guard data.count > 0 else { return nil }
return data
}
}
And for the sake of completeness, this is my Data to hex string routine:
extension Data {
/// Hexadecimal string representation of `Data` object.
var hexadecimal: String {
return map { String(format: "%02x", $0) }
.joined()
}
}
Note, as shown in the above, I generally only convert between hexadecimal representations and NSData instances (because if the information could have been represented as a string you probably wouldn't have created a hexadecimal representation in the first place). But your original question wanted to convert between hexadecimal representations and String objects, and that might look like so:
extension String {
/// Create `String` representation of `Data` created from hexadecimal string representation
///
/// This takes a hexadecimal representation and creates a String object from that. Note, if the string has any spaces, those are removed. Also if the string started with a `<` or ended with a `>`, those are removed, too.
///
/// For example,
///
/// String(hexadecimal: "<666f6f>")
///
/// is
///
/// Optional("foo")
///
/// - returns: `String` represented by this hexadecimal string.
init?(hexadecimal string: String, encoding: String.Encoding = .utf8) {
guard let data = string.hexadecimal() else {
return nil
}
self.init(data: data, encoding: encoding)
}
/// Create hexadecimal string representation of `String` object.
///
/// For example,
///
/// "foo".hexadecimalString()
///
/// is
///
/// Optional("666f6f")
///
/// - parameter encoding: The `String.Encoding` that indicates how the string should be converted to `Data` before performing the hexadecimal conversion.
///
/// - returns: `String` representation of this String object.
func hexadecimalString(encoding: String.Encoding = .utf8) -> String? {
return data(using: encoding)?
.hexadecimal
}
}
You could then use the above like so:
let hexString = "68656c6c 6f2c2077 6f726c64"
print(String(hexadecimal: hexString))
Or,
let originalString = "hello, world"
print(originalString.hexadecimalString())
For permutations of the above for earlier Swift versions, see the revision history of this question.
convert hex string to data and string:
Swift1
func dataWithHexString(hex: String) -> NSData {
var hex = hex
let data = NSMutableData()
while(countElements(hex) > 0) {
var c: String = hex.substringToIndex(advance(hex.startIndex, 2))
hex = hex.substringFromIndex(advance(hex.startIndex, 2))
var ch: UInt32 = 0
NSScanner(string: c).scanHexInt(&ch)
data.appendBytes(&ch, length: 1)
}
return data
}
use:
let data = dataWithHexString("68656c6c6f2c20776f726c64") // <68656c6c 6f2c2077 6f726c64>
if let string = NSString(data: data, encoding: 1) {
print(string) // hello, world
}
Swift2
func dataWithHexString(hex: String) -> NSData {
var hex = hex
let data = NSMutableData()
while(hex.characters.count > 0) {
let c: String = hex.substringToIndex(hex.startIndex.advancedBy(2))
hex = hex.substringFromIndex(hex.startIndex.advancedBy(2))
var ch: UInt32 = 0
NSScanner(string: c).scanHexInt(&ch)
data.appendBytes(&ch, length: 1)
}
return data
}
use:
let data = dataWithHexString("68656c6c6f2c20776f726c64") // <68656c6c 6f2c2077 6f726c64>
if let string = String(data: data, encoding: NSUTF8StringEncoding) {
print(string) //"hello, world"
}
Swift3
func dataWithHexString(hex: String) -> Data {
var hex = hex
var data = Data()
while(hex.characters.count > 0) {
let c: String = hex.substring(to: hex.index(hex.startIndex, offsetBy: 2))
hex = hex.substring(from: hex.index(hex.startIndex, offsetBy: 2))
var ch: UInt32 = 0
Scanner(string: c).scanHexInt32(&ch)
var char = UInt8(ch)
data.append(&char, count: 1)
}
return data
}
use:
let data = dataWithHexString(hex: "68656c6c6f2c20776f726c64") // <68656c6c 6f2c2077 6f726c64>
let string = String(data: data, encoding: .utf8) // "hello, world"
Swift4
func dataWithHexString(hex: String) -> Data {
var hex = hex
var data = Data()
while(hex.count > 0) {
let subIndex = hex.index(hex.startIndex, offsetBy: 2)
let c = String(hex[..<subIndex])
hex = String(hex[subIndex...])
var ch: UInt32 = 0
Scanner(string: c).scanHexInt32(&ch)
var char = UInt8(ch)
data.append(&char, count: 1)
}
return data
}
use:
let data = dataWithHexString(hex: "68656c6c6f2c20776f726c64") // <68656c6c 6f2c2077 6f726c64>
let string = String(data: data, encoding: .utf8) // "hello, world"
Swift 4 & Swift 5 implementation:
init?(hexString: String) {
let len = hexString.count / 2
var data = Data(capacity: len)
var i = hexString.startIndex
for _ in 0..<len {
let j = hexString.index(i, offsetBy: 2)
let bytes = hexString[i..<j]
if var num = UInt8(bytes, radix: 16) {
data.append(&num, count: 1)
} else {
return nil
}
i = j
}
self = data
}
Usage:
let data = Data(hexString: "0a1b3c4d")
Swift 5
extension Data {
init?(hex: String) {
guard hex.count.isMultiple(of: 2) else {
return nil
}
let chars = hex.map { $0 }
let bytes = stride(from: 0, to: chars.count, by: 2)
.map { String(chars[$0]) + String(chars[$0 + 1]) }
.compactMap { UInt8($0, radix: 16) }
guard hex.count / bytes.count == 2 else { return nil }
self.init(bytes)
}
}
Here is my Swift 5 way to do it:
does take care of "0x" prefixes
use subscript instead of allocated Array(), no C style [i+1] too
add .hexadecimal to String.data(using encoding:) -> Data?
.
String Extension:
extension String {
enum ExtendedEncoding {
case hexadecimal
}
func data(using encoding:ExtendedEncoding) -> Data? {
let hexStr = self.dropFirst(self.hasPrefix("0x") ? 2 : 0)
guard hexStr.count % 2 == 0 else { return nil }
var newData = Data(capacity: hexStr.count/2)
var indexIsEven = true
for i in hexStr.indices {
if indexIsEven {
let byteRange = i...hexStr.index(after: i)
guard let byte = UInt8(hexStr[byteRange], radix: 16) else { return nil }
newData.append(byte)
}
indexIsEven.toggle()
}
return newData
}
}
Usage:
"5413".data(using: .hexadecimal)
"0x1234FF".data(using: .hexadecimal)
Tests:
extension Data {
var bytes:[UInt8] { // fancy pretty call: myData.bytes -> [UInt8]
return [UInt8](self)
}
// Could make a more optimized one~
func hexa(prefixed isPrefixed:Bool = true) -> String {
return self.bytes.reduce(isPrefixed ? "0x" : "") { $0 + String(format: "%02X", $1) }
}
}
print("000204ff5400".data(using: .hexadecimal)?.hexa() ?? "failed") // OK
print("0x000204ff5400".data(using: .hexadecimal)?.hexa() ?? "failed") // OK
print("541".data(using: .hexadecimal)?.hexa() ?? "failed") // fails
print("5413".data(using: .hexadecimal)?.hexa() ?? "failed") // OK
Here's a simple solution I settled on:
extension NSData {
public convenience init(hexString: String) {
var index = hexString.startIndex
var bytes: [UInt8] = []
repeat {
bytes.append(hexString[index...index.advancedBy(1)].withCString {
return UInt8(strtoul($0, nil, 16))
})
index = index.advancedBy(2)
} while index.distanceTo(hexString.endIndex) != 0
self.init(bytes: &bytes, length: bytes.count)
}
}
Usage:
let data = NSData(hexString: "b8dfb080bc33fb564249e34252bf143d88fc018f")
Output:
print(data)
>>> <b8dfb080 bc33fb56 4249e342 52bf143d 88fc018f>
Update 6/29/2016
I updated the initializer to handle malformed data (i.e., invalid characters or odd number of characters).
public convenience init?(hexString: String, force: Bool) {
let characterSet = NSCharacterSet(charactersInString: "0123456789abcdefABCDEF")
for scalar in hexString.unicodeScalars {
if characterSet.characterIsMember(UInt16(scalar.value)) {
hexString.append(scalar)
}
else if !force {
return nil
}
}
if hexString.characters.count % 2 == 1 {
if force {
hexString = "0" + hexString
}
else {
return nil
}
}
var index = hexString.startIndex
var bytes: [UInt8] = []
repeat {
bytes.append(hexString[index...index.advancedBy(1)].withCString {
return UInt8(strtoul($0, nil, 16))
})
index = index.advancedBy(2)
} while index.distanceTo(hexString.endIndex) != 0
self.init(bytes: &bytes, length: bytes.count)
}
Here is my take on converting hexadecimal string to Data using Swift 4:
extension Data {
private static let hexRegex = try! NSRegularExpression(pattern: "^([a-fA-F0-9][a-fA-F0-9])*$", options: [])
init?(hexString: String) {
if Data.hexRegex.matches(in: hexString, range: NSMakeRange(0, hexString.count)).isEmpty {
return nil // does not look like a hexadecimal string
}
let chars = Array(hexString)
let bytes: [UInt8] =
stride(from: 0, to: chars.count, by: 2)
.map {UInt8(String([chars[$0], chars[$0+1]]), radix: 16)}
.compactMap{$0}
self = Data(bytes)
}
var hexString: String {
return map { String(format: "%02hhx", $0) }.joined()
}
}
(I threw in a small feature for converting back to hex string I found in this answer)
And here is how you would use it:
let data = Data(hexString: "cafecafe")
print(data?.hexString) // will print Optional("cafecafe")
One more solution that is simple to follow and leverages swifts built-in hex parsing
func convertHexToBytes(_ str: String) -> Data? {
let values = str.compactMap { $0.hexDigitValue } // map char to value of 0-15 or nil
if values.count == str.count && values.count % 2 == 0 {
var data = Data()
for x in stride(from: 0, to: values.count, by: 2) {
let byte = (values[x] << 4) + values[x+1] // concat high and low bits
data.append(UInt8(byte))
}
return data
}
return nil
}
let good = "e01AFd"
let bad = "e0671"
let ugly = "GT40"
print("\(convertHexToBytes(good))") // Optional(3 bytes)
print("\(convertHexToBytes(bad))") // nil
print("\(convertHexToBytes(ugly))") // nil
The code worked for me in Swift 3.0.2.
extension String {
/// Expanded encoding
///
/// - bytesHexLiteral: Hex string of bytes
/// - base64: Base64 string
enum ExpandedEncoding {
/// Hex string of bytes
case bytesHexLiteral
/// Base64 string
case base64
}
/// Convert to `Data` with expanded encoding
///
/// - Parameter encoding: Expanded encoding
/// - Returns: data
func data(using encoding: ExpandedEncoding) -> Data? {
switch encoding {
case .bytesHexLiteral:
guard self.characters.count % 2 == 0 else { return nil }
var data = Data()
var byteLiteral = ""
for (index, character) in self.characters.enumerated() {
if index % 2 == 0 {
byteLiteral = String(character)
} else {
byteLiteral.append(character)
guard let byte = UInt8(byteLiteral, radix: 16) else { return nil }
data.append(byte)
}
}
return data
case .base64:
return Data(base64Encoded: self)
}
}
}
Swift 5
With support iOS 13 and iOS2...iOS12.
extension String {
var hex: Data? {
var value = self
var data = Data()
while value.count > 0 {
let subIndex = value.index(value.startIndex, offsetBy: 2)
let c = String(value[..<subIndex])
value = String(value[subIndex...])
var char: UInt8
if #available(iOS 13.0, *) {
guard let int = Scanner(string: c).scanInt32(representation: .hexadecimal) else { return nil }
char = UInt8(int)
} else {
var int: UInt32 = 0
Scanner(string: c).scanHexInt32(&int)
char = UInt8(int)
}
data.append(&char, count: 1)
}
return data
}
}
Swift 5
There is a compact implementation of initialize Data instance from hex string using a regular expression. It searches hex numbers inside a string and combine them to a result data so that it can support different formats of hex representations:
extension Data {
private static let regex = try! NSRegularExpression(pattern: "([0-9a-fA-F]{2})", options: [])
/// Create instance from string with hex numbers.
init(from: String) {
let range = NSRange(location: 0, length: from.utf16.count)
let bytes = Self.regex.matches(in: from, options: [], range: range)
.compactMap { Range($0.range(at: 1), in: from) }
.compactMap { UInt8(from[$0], radix: 16) }
self.init(bytes)
}
/// Hex string representation of data.
var hex: String {
map { String($0, radix: 16) }.joined()
}
}
Examples:
let data = Data(from: "0x11223344aabbccdd")
print(data.hex) // Prints "11223344aabbccdd"
let data2 = Data(from: "11223344aabbccdd")
print(data2.hex) // Prints "11223344aabbccdd"
let data3 = Data(from: "11223344 aabbccdd")
print(data3.hex) // Prints "11223344aabbccdd"
let data4 = Data(from: "11223344 AABBCCDD")
print(data4.hex) // Prints "11223344aabbccdd"
let data5 = Data(from: "Hex: 0x11223344AABBCCDD")
print(data5.hex) // Prints "11223344aabbccdd"
let data6 = Data(from: "word[0]=11223344 word[1]=AABBCCDD")
print(data6.hex) // Prints "11223344aabbccdd"
let data7 = Data(from: "No hex")
print(data7.hex) // Prints ""
Handles prefixes
Ignores invalid characters and incomplete bytes
Uses Swift built in hex character parsing
Doesn't use subscripts
extension Data {
init(hexString: String) {
self = hexString
.dropFirst(hexString.hasPrefix("0x") ? 2 : 0)
.compactMap { $0.hexDigitValue.map { UInt8($0) } }
.reduce(into: (data: Data(capacity: hexString.count / 2), byte: nil as UInt8?)) { partialResult, nibble in
if let p = partialResult.byte {
partialResult.data.append(p + nibble)
partialResult.byte = nil
} else {
partialResult.byte = nibble << 4
}
}.data
}
}
Supposing your string is even size, you can use this to convert to hexadecimal and save it to Data:
Swift 5.2
func hex(from string: String) -> Data {
.init(stride(from: 0, to: string.count, by: 2).map {
string[string.index(string.startIndex, offsetBy: $0) ... string.index(string.startIndex, offsetBy: $0 + 1)]
}.map {
UInt8($0, radix: 16)!
})
}