Get nth character of a string in Swift programming language
SwiftStringCollectionsCharacterSubscriptSwift Problem Overview
How can I get the nth character of a string? I tried bracket([]) accessor with no luck.
var string = "Hello, world!"
var firstChar = string[0] // Throws error
> ERROR: 'subscript' is unavailable: cannot subscript String with an Int, see the documentation comment for discussion
Swift Solutions
Solution 1 - Swift
Attention: Please see Leo Dabus' answer for a proper implementation for Swift 4 and Swift 5.
Swift 4 or later
The Substring type was introduced in Swift 4 to make substrings
faster and more efficient by sharing storage with the original string, so that's what the subscript functions should return.
Try it out here
extension StringProtocol {
subscript(offset: Int) -> Character { self[index(startIndex, offsetBy: offset)] }
subscript(range: Range<Int>) -> SubSequence {
let startIndex = index(self.startIndex, offsetBy: range.lowerBound)
return self[startIndex..<index(startIndex, offsetBy: range.count)]
}
subscript(range: ClosedRange<Int>) -> SubSequence {
let startIndex = index(self.startIndex, offsetBy: range.lowerBound)
return self[startIndex..<index(startIndex, offsetBy: range.count)]
}
subscript(range: PartialRangeFrom<Int>) -> SubSequence { self[index(startIndex, offsetBy: range.lowerBound)...] }
subscript(range: PartialRangeThrough<Int>) -> SubSequence { self[...index(startIndex, offsetBy: range.upperBound)] }
subscript(range: PartialRangeUpTo<Int>) -> SubSequence { self[..<index(startIndex, offsetBy: range.upperBound)] }
}
To convert the Substring into a String, you can simply
do String(string[0..2]), but you should only do that if
you plan to keep the substring around. Otherwise, it's more
efficient to keep it a Substring.
It would be great if someone could figure out a good way to merge
these two extensions into one. I tried extending Note: This answer has been already edited, it is properly implemented and now works for substrings as well. Just make sure to use a valid range to avoid crashing when subscripting your StringProtocol type. For subscripting with a range that won't crash with out of range values you can use this implementationStringProtocol
without success, because the index method does not exist there.
Why is this not built-in?
The error message says "see the documentation comment for discussion". Apple provides the following explanation in the file UnavailableStringAPIs.swift:
> Subscripting strings with integers is not available.
>
> The concept of "the ith character in a string" has
> different interpretations in different libraries and system
> components. The correct interpretation should be selected
> according to the use case and the APIs involved, so String
> cannot be subscripted with an integer.
>
> Swift provides several different ways to access the character
> data stored inside strings.
>
> - String.utf8 is a collection of UTF-8 code units in the
> string. Use this API when converting the string to UTF-8.
> Most POSIX APIs process strings in terms of UTF-8 code units.
>
> - String.utf16 is a collection of UTF-16 code units in
> string. Most Cocoa and Cocoa touch APIs process strings in
> terms of UTF-16 code units. For example, instances of
> NSRange used with NSAttributedString and
> NSRegularExpression store substring offsets and lengths in
> terms of UTF-16 code units.
>
> - String.unicodeScalars is a collection of Unicode scalars.
> Use this API when you are performing low-level manipulation
> of character data.
>
> - String.characters is a collection of extended grapheme
> clusters, which are an approximation of user-perceived
> characters.
>
> Note that when processing strings that contain human-readable text,
> character-by-character processing should be avoided to the largest extent
> possible. Use high-level locale-sensitive Unicode algorithms instead, for example,
> String.localizedStandardCompare(),
> String.localizedLowercaseString,
> String.localizedStandardRangeOfString() etc.
Solution 2 - Swift
Swift 5.2
let str = "abcdef"
str[1 ..< 3] // returns "bc"
str[5] // returns "f"
str[80] // returns ""
str.substring(fromIndex: 3) // returns "def"
str.substring(toIndex: str.length - 2) // returns "abcd"
You will need to add this String extension to your project (it's fully tested):
extension String {
var length: Int {
return count
}
subscript (i: Int) -> String {
return self[i ..< i + 1]
}
func substring(fromIndex: Int) -> String {
return self[min(fromIndex, length) ..< length]
}
func substring(toIndex: Int) -> String {
return self[0 ..< max(0, toIndex)]
}
subscript (r: Range<Int>) -> String {
let range = Range(uncheckedBounds: (lower: max(0, min(length, r.lowerBound)),
upper: min(length, max(0, r.upperBound))))
let start = index(startIndex, offsetBy: range.lowerBound)
let end = index(start, offsetBy: range.upperBound - range.lowerBound)
return String(self[start ..< end])
}
}
Even though Swift always had out of the box solution to this problem (without String extension, which I provided below), I still would strongly recommend using the extension. Why? Because it saved me tens of hours of painful migration from early versions of Swift, where String's syntax was changing almost every release, but all I needed to do was to update the extension's implementation as opposed to refactoring the entire project. Make your choice.
let str = "Hello, world!"
let index = str.index(str.startIndex, offsetBy: 4)
str[index] // returns Character 'o'
let endIndex = str.index(str.endIndex, offsetBy:-2)
str[index ..< endIndex] // returns String "o, worl"
String(str.suffix(from: index)) // returns String "o, world!"
String(str.prefix(upTo: index)) // returns String "Hell"
Solution 3 - Swift
I just came up with this neat workaround
var firstChar = Array(string)[0]
Solution 4 - Swift
Xcode 11 β’ Swift 5.1
You can extend StringProtocol to make the subscript available also to the substrings:
extension StringProtocol {
subscript(_ offset: Int) -> Element { self[index(startIndex, offsetBy: offset)] }
subscript(_ range: Range<Int>) -> SubSequence { prefix(range.lowerBound+range.count).suffix(range.count) }
subscript(_ range: ClosedRange<Int>) -> SubSequence { prefix(range.lowerBound+range.count).suffix(range.count) }
subscript(_ range: PartialRangeThrough<Int>) -> SubSequence { prefix(range.upperBound.advanced(by: 1)) }
subscript(_ range: PartialRangeUpTo<Int>) -> SubSequence { prefix(range.upperBound) }
subscript(_ range: PartialRangeFrom<Int>) -> SubSequence { suffix(Swift.max(0, count-range.lowerBound)) }
}
extension LosslessStringConvertible {
var string: String { .init(self) }
}
extension BidirectionalCollection {
subscript(safe offset: Int) -> Element? {
guard !isEmpty, let i = index(startIndex, offsetBy: offset, limitedBy: index(before: endIndex)) else { return nil }
return self[i]
}
}
Testing
let test = "Hello USA πΊπΈ!!! Hello Brazil π§π·!!!"
test[safe: 10] // "πΊπΈ"
test[11] // "!"
test[10...] // "πΊπΈ!!! Hello Brazil π§π·!!!"
test[10..<12] // "πΊπΈ!"
test[10...12] // "πΊπΈ!!"
test[...10] // "Hello USA πΊπΈ"
test[..<10] // "Hello USA "
test.first // "H"
test.last // "!"
// Subscripting the Substring
test[...][...3] // "Hell"
// Note that they all return a Substring of the original String.
// To create a new String from a substring
test[10...].string // "πΊπΈ!!! Hello Brazil π§π·!!!"
Solution 5 - Swift
No indexing using integers, only using String.Index. Mostly with linear complexity. You can also create ranges from String.Index and get substrings using them.
Swift 3.0
let firstChar = someString[someString.startIndex]
let lastChar = someString[someString.index(before: someString.endIndex)]
let charAtIndex = someString[someString.index(someString.startIndex, offsetBy: 10)]
let range = someString.startIndex..<someString.index(someString.startIndex, offsetBy: 10)
let substring = someString[range]
Swift 2.x
let firstChar = someString[someString.startIndex]
let lastChar = someString[someString.endIndex.predecessor()]
let charAtIndex = someString[someString.startIndex.advanceBy(10)]
let range = someString.startIndex..<someString.startIndex.advanceBy(10)
let subtring = someString[range]
Note that you can't ever use an index (or range) created from one string to another string
let index10 = someString.startIndex.advanceBy(10)
//will compile
//sometimes it will work but sometimes it will crash or result in undefined behaviour
let charFromAnotherString = anotherString[index10]
Solution 6 - Swift
Swift 4
let str = "My String"
String at index
let index = str.index(str.startIndex, offsetBy: 3)
String(str[index]) // "S"
Substring
let startIndex = str.index(str.startIndex, offsetBy: 3)
let endIndex = str.index(str.startIndex, offsetBy: 7)
String(str[startIndex...endIndex]) // "Strin"
First n chars
let startIndex = str.index(str.startIndex, offsetBy: 3)
String(str[..<startIndex]) // "My "
Last n chars
let startIndex = str.index(str.startIndex, offsetBy: 3)
String(str[startIndex...]) // "String"
Swift 2 and 3
str = "My String"
**String At Index **
Swift 2
let charAtIndex = String(str[str.startIndex.advancedBy(3)]) // charAtIndex = "S"
Swift 3
str[str.index(str.startIndex, offsetBy: 3)]
SubString fromIndex toIndex
Swift 2
let subStr = str[str.startIndex.advancedBy(3)...str.startIndex.advancedBy(7)] // subStr = "Strin"
Swift 3
str[str.index(str.startIndex, offsetBy: 3)...str.index(str.startIndex, offsetBy: 7)]
First n chars
let first2Chars = String(str.characters.prefix(2)) // first2Chars = "My"
Last n chars
let last3Chars = String(str.characters.suffix(3)) // last3Chars = "ing"
Solution 7 - Swift
Swift 5.3
I think this is very elegant. Kudos at Paul Hudson of "Hacking with Swift" for this solution:
@available (macOS 10.15, * )
extension String {
subscript(idx: Int) -> String {
String(self[index(startIndex, offsetBy: idx)])
}
}
Then to get one character out of the String you simply do:
var string = "Hello, world!"
var firstChar = string[0] // No error, returns "H" as a String
NB: I just wanted to add, this will return a String as pointed out in the comments. I think it might be unexpected for Swift users, but often I need a String to use in my code straight away and not a Character type, so it does simplify my code a little bit avoiding a conversion from Character to String later.
Solution 8 - Swift
If you see Cannot subscript a value of type 'String'... use this extension:
Swift 3
extension String {
subscript (i: Int) -> Character {
return self[self.characters.index(self.startIndex, offsetBy: i)]
}
subscript (i: Int) -> String {
return String(self[i] as Character)
}
subscript (r: Range<Int>) -> String {
let start = index(startIndex, offsetBy: r.lowerBound)
let end = index(startIndex, offsetBy: r.upperBound)
return self[start..<end]
}
subscript (r: ClosedRange<Int>) -> String {
let start = index(startIndex, offsetBy: r.lowerBound)
let end = index(startIndex, offsetBy: r.upperBound)
return self[start...end]
}
}
Swift 2.3
extension String {
subscript(integerIndex: Int) -> Character {
let index = advance(startIndex, integerIndex)
return self[index]
}
subscript(integerRange: Range<Int>) -> String {
let start = advance(startIndex, integerRange.startIndex)
let end = advance(startIndex, integerRange.endIndex)
let range = start..<end
return self[range]
}
}
Solution 9 - Swift
Swift 2.0 as of Xcode 7 GM Seed
var text = "Hello, world!"
let firstChar = text[text.startIndex.advancedBy(0)] // "H"
For the nth character, replace 0 with n-1.
Edit: Swift 3.0
text[text.index(text.startIndex, offsetBy: 0)]
n.b. there are simpler ways of grabbing certain characters in the string
e.g. let firstChar = text.characters.first
Solution 10 - Swift
Swift 2.2 Solution:
The following extension works in Xcode 7, this is a combination of this solution and Swift 2.0 syntax conversion.
extension String {
subscript(integerIndex: Int) -> Character {
let index = startIndex.advancedBy(integerIndex)
return self[index]
}
subscript(integerRange: Range<Int>) -> String {
let start = startIndex.advancedBy(integerRange.startIndex)
let end = startIndex.advancedBy(integerRange.endIndex)
let range = start..<end
return self[range]
}
}
Solution 11 - Swift
The swift string class does not provide the ability to get a character at a specific index because of its native support for UTF characters. The variable length of a UTF character in memory makes jumping directly to a character impossible. That means you have to manually loop over the string each time.
You can extend String to provide a method that will loop through the characters until your desired index
extension String {
func characterAtIndex(index: Int) -> Character? {
var cur = 0
for char in self {
if cur == index {
return char
}
cur++
}
return nil
}
}
myString.characterAtIndex(0)!
Solution 12 - Swift
As an aside note, there are a few functions applyable directly to the Character-chain representation of a String, like this:
var string = "Hello, playground"
let firstCharacter = string.characters.first // returns "H"
let lastCharacter = string.characters.last // returns "d"
The result is of type Character, but you can cast it to a String.
Or this:
let reversedString = String(string.characters.reverse())
// returns "dnuorgyalp ,olleH"
:-)
Solution 13 - Swift
Swift 4
String(Array(stringToIndex)[index])
This is probably the best way of solving this problem one-time. You probably want to cast the String as an array first, and then cast the result as a String again. Otherwise, a Character will be returned instead of a String.
Example String(Array("HelloThere")[1]) will return "e" as a String.
(Array("HelloThere")[1] will return "e" as a Character.
Swift does not allow Strings to be indexed like arrays, but this gets the job done, brute-force style.
Solution 14 - Swift
You can do it by convert String into Array and get it by specific index using subscript as below
var str = "Hello"
let s = Array(str)[2]
print(s)
Solution 15 - Swift
My very simple solution:
Swift 4.1:
let myString = "Test string"
let index = 0
let firstCharacter = myString[String.Index(encodedOffset: index)]
Swift 5.1:
let firstCharacter = myString[String.Index.init(utf16Offset: index, in: myString)]
Solution 16 - Swift
I just had the same issue. Simply do this:
var aString: String = "test"
var aChar:unichar = (aString as NSString).characterAtIndex(0)
Solution 17 - Swift
Swift3
You can use subscript syntax to access the Character at a particular String index.
let greeting = "Guten Tag!"
let index = greeting.index(greeting.startIndex, offsetBy: 7)
greeting[index] // a
or we can do a String Extension in Swift 4
extension String {
func getCharAtIndex(_ index: Int) -> Character {
return self[self.index(self.startIndex, offsetBy: index)]
}
}
USAGE:
let foo = "ABC123"
foo.getCharAtIndex(2) //C
Solution 18 - Swift
By now, subscript(_:) is unavailable. As well as we can't do this
str[0]
with string.We have to provide "String.Index" But, how can we give our own index number in this way, instead we can use,
string[str.index(str.startIndex, offsetBy: 0)]
Solution 19 - Swift
In Swift 5 without extension to the String :
var str = "ABCDEFGH"
for char in str {
if(char == "C") { }
}
Above Swift code as same as that Java code :
int n = 8;
var str = "ABCDEFGH"
for (int i=0; i<n; i++) {
if (str.charAt(i) == 'C') { }
}
Solution 20 - Swift
Swift 5.1.3:
Add a String extension:
extension String {
func stringAt(_ i: Int) -> String {
return String(Array(self)[i])
}
func charAt(_ i: Int) -> Character {
return Array(self)[i]
}
}
let str = "Teja Kumar"
let str1: String = str.stringAt(2) //"j"
let str2: Character = str.charAt(5) //"k"
Solution 21 - Swift
My solution is in one line, supposing cadena is the string and 4 is the nth position that you want:
let character = cadena[advance(cadena.startIndex, 4)]
Simple... I suppose Swift will include more things about substrings in future versions.
Solution 22 - Swift
Swift 3: another solution (tested in playground)
extension String {
func substr(_ start:Int, length:Int=0) -> String? {
guard start > -1 else {
return nil
}
let count = self.characters.count - 1
guard start <= count else {
return nil
}
let startOffset = max(0, start)
let endOffset = length > 0 ? min(count, startOffset + length - 1) : count
return self[self.index(self.startIndex, offsetBy: startOffset)...self.index(self.startIndex, offsetBy: endOffset)]
}
}
Usage:
let txt = "12345"
txt.substr(-1) //nil
txt.substr(0) //"12345"
txt.substr(0, length: 0) //"12345"
txt.substr(1) //"2345"
txt.substr(2) //"345"
txt.substr(3) //"45"
txt.substr(4) //"5"
txt.substr(6) //nil
txt.substr(0, length: 1) //"1"
txt.substr(1, length: 1) //"2"
txt.substr(2, length: 1) //"3"
txt.substr(3, length: 1) //"4"
txt.substr(3, length: 2) //"45"
txt.substr(3, length: 3) //"45"
txt.substr(4, length: 1) //"5"
txt.substr(4, length: 2) //"5"
txt.substr(5, length: 1) //nil
txt.substr(5, length: -1) //nil
txt.substr(-1, length: -1) //nil
Solution 23 - Swift
Swift 4.2 or later
Range and partial range subscripting using String's indices property
As variation of @LeoDabus nice answer, we may add an additional extension to DefaultIndices with the purpose of allowing us to fall back on the indices property of String when implementing the custom subscripts (by Int specialized ranges and partial ranges) for the latter.
extension DefaultIndices {
subscript(at: Int) -> Elements.Index { index(startIndex, offsetBy: at) }
}
// Moving the index(_:offsetBy:) to an extension yields slightly
// briefer implementations for these String extensions.
extension String {
subscript(range: Range<Int>) -> SubSequence {
let start = indices[range.lowerBound]
return self[start..<indices[start...][range.count]]
}
subscript(range: ClosedRange<Int>) -> SubSequence {
let start = indices[range.lowerBound]
return self[start...indices[start...][range.count]]
}
subscript(range: PartialRangeFrom<Int>) -> SubSequence {
self[indices[range.lowerBound]...]
}
subscript(range: PartialRangeThrough<Int>) -> SubSequence {
self[...indices[range.upperBound]]
}
subscript(range: PartialRangeUpTo<Int>) -> SubSequence {
self[..<indices[range.upperBound]]
}
}
let str = "foo bar baz bax"
print(str[4..<6]) // "ba"
print(str[4...6]) // "bar"
print(str[4...]) // "bar baz bax"
print(str[...6]) // "foo bar"
print(str[..<6]) // "foo ba"
Thanks @LeoDabus for the pointing me in the direction of using the indices property as an(other) alternative to String subscripting!
Solution 24 - Swift
Best way which worked for me is:
var firstName = "Olivia"
var lastName = "Pope"
var nameInitials.text = "\(firstName.prefix(1))" + "\ (lastName.prefix(1))"
Output:"OP"
Solution 25 - Swift
We have subscript which will very useful here
But String subscript will take param as String.Index so most of the people gets confuse here how to pass String.Index to get details how to form String.Index as per our requirement please look at below documentation Apple Documentation
Here i have created one extension method to get nth character in string
extension String {
subscript(i: Int) -> String {
return i < count ? String(self[index(startIndex, offsetBy: i)]) : ""
}
}
Usage
let name = "Narayana Rao"
print(name[11]) //o
print(name[1]) //a
print(name[0]) //N
print(name[30]) //""
if you pass index which is out of bounds of String count it will return empty String
Solution 26 - Swift
Update for swift 2.0 subString
public extension String {
public subscript (i: Int) -> String {
return self.substringWithRange(self.startIndex..<self.startIndex.advancedBy(i + 1))
}
public subscript (r: Range<Int>) -> String {
get {
return self.substringWithRange(self.startIndex.advancedBy(r.startIndex)..<self.startIndex.advancedBy(r.endIndex))
}
}
}
Solution 27 - Swift
I think that a fast answer for get the first character could be:
let firstCharacter = aString[aString.startIndex]
It's so much elegant and performance than:
let firstCharacter = Array(aString.characters).first
But.. if you want manipulate and do more operations with strings you could think create an extension..here is one extension with this approach, it's quite similar to that already posted here:
extension String {
var length : Int {
return self.characters.count
}
subscript(integerIndex: Int) -> Character {
let index = startIndex.advancedBy(integerIndex)
return self[index]
}
subscript(integerRange: Range<Int>) -> String {
let start = startIndex.advancedBy(integerRange.startIndex)
let end = startIndex.advancedBy(integerRange.endIndex)
let range = start..<end
return self[range]
}
}
BUT IT'S A TERRIBLE IDEA!!
The extension below is horribly inefficient. Every time a string is accessed with an integer, an O(n) function to advance its starting index is run. Running a linear loop inside another linear loop means this for loop is accidentally O(n2) β as the length of the string increases, the time this loop takes increases quadratically.
Instead of doing that you could use the characters's string collection.
Solution 28 - Swift
Swift 3
extension String {
public func charAt(_ i: Int) -> Character {
return self[self.characters.index(self.startIndex, offsetBy: i)]
}
public subscript (i: Int) -> String {
return String(self.charAt(i) as Character)
}
public subscript (r: Range<Int>) -> String {
return substring(with: self.characters.index(self.startIndex, offsetBy: r.lowerBound)..<self.characters.index(self.startIndex, offsetBy: r.upperBound))
}
public subscript (r: CountableClosedRange<Int>) -> String {
return substring(with: self.characters.index(self.startIndex, offsetBy: r.lowerBound)..<self.characters.index(self.startIndex, offsetBy: r.upperBound))
}
}
Usage
let str = "Hello World"
let sub = str[0...4]
Solution 29 - Swift
Here's an extension you can use, working with Swift 3.1. A single index will return a Character, which seems intuitive when indexing a String, and a Range will return a String.
extension String {
subscript (i: Int) -> Character {
return Array(self.characters)[i]
}
subscript (r: CountableClosedRange<Int>) -> String {
return String(Array(self.characters)[r])
}
subscript (r: CountableRange<Int>) -> String {
return self[r.lowerBound...r.upperBound-1]
}
}
Some examples of the extension in action:
let string = "Hello"
let c1 = string[1] // Character "e"
let c2 = string[-1] // fatal error: Index out of range
let r1 = string[1..<4] // String "ell"
let r2 = string[1...4] // String "ello"
let r3 = string[1...5] // fatal error: Array index is out of range
n.b. You could add an additional method to the above extension to return a String with a single character if wanted:
subscript (i: Int) -> String {
return String(self[i])
}
Note that then you would have to explicitly specify the type you wanted when indexing the string:
let c: Character = string[3] // Character "l"
let s: String = string[0] // String "H"
Solution 30 - Swift
Get & Set Subscript (String & Substring) - Swift 4.2
Swift 4.2, Xcode 10
I based my answer off of @alecarlson's answer.
The only big difference is you can get a Substring or a String returned (and in some cases, a single Character). You can also get and set the subscript.
Lastly, mine is a bit more cumbersome and longer than @alecarlson's answer and as such, I suggest you put it in a source file.
Extension:
public extension String {
public subscript (i: Int) -> Character {
get {
return self[index(startIndex, offsetBy: i)]
}
set (c) {
let n = index(startIndex, offsetBy: i)
replaceSubrange(n...n, with: "\(c)")
}
}
public subscript (bounds: CountableRange<Int>) -> Substring {
get {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[start ..< end]
}
set (s) {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
replaceSubrange(start ..< end, with: s)
}
}
public subscript (bounds: CountableClosedRange<Int>) -> Substring {
get {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[start ... end]
}
set (s) {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
replaceSubrange(start ... end, with: s)
}
}
public subscript (bounds: CountablePartialRangeFrom<Int>) -> Substring {
get {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(endIndex, offsetBy: -1)
return self[start ... end]
}
set (s) {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(endIndex, offsetBy: -1)
replaceSubrange(start ... end, with: s)
}
}
public subscript (bounds: PartialRangeThrough<Int>) -> Substring {
get {
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[startIndex ... end]
}
set (s) {
let end = index(startIndex, offsetBy: bounds.upperBound)
replaceSubrange(startIndex ... end, with: s)
}
}
public subscript (bounds: PartialRangeUpTo<Int>) -> Substring {
get {
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[startIndex ..< end]
}
set (s) {
let end = index(startIndex, offsetBy: bounds.upperBound)
replaceSubrange(startIndex ..< end, with: s)
}
}
public subscript (i: Int) -> String {
get {
return "\(self[index(startIndex, offsetBy: i)])"
}
set (c) {
let n = index(startIndex, offsetBy: i)
self.replaceSubrange(n...n, with: "\(c)")
}
}
public subscript (bounds: CountableRange<Int>) -> String {
get {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
return "\(self[start ..< end])"
}
set (s) {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
replaceSubrange(start ..< end, with: s)
}
}
public subscript (bounds: CountableClosedRange<Int>) -> String {
get {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
return "\(self[start ... end])"
}
set (s) {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
replaceSubrange(start ... end, with: s)
}
}
public subscript (bounds: CountablePartialRangeFrom<Int>) -> String {
get {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(endIndex, offsetBy: -1)
return "\(self[start ... end])"
}
set (s) {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(endIndex, offsetBy: -1)
replaceSubrange(start ... end, with: s)
}
}
public subscript (bounds: PartialRangeThrough<Int>) -> String {
get {
let end = index(startIndex, offsetBy: bounds.upperBound)
return "\(self[startIndex ... end])"
}
set (s) {
let end = index(startIndex, offsetBy: bounds.upperBound)
replaceSubrange(startIndex ... end, with: s)
}
}
public subscript (bounds: PartialRangeUpTo<Int>) -> String {
get {
let end = index(startIndex, offsetBy: bounds.upperBound)
return "\(self[startIndex ..< end])"
}
set (s) {
let end = index(startIndex, offsetBy: bounds.upperBound)
replaceSubrange(startIndex ..< end, with: s)
}
}
public subscript (i: Int) -> Substring {
get {
return Substring("\(self[index(startIndex, offsetBy: i)])")
}
set (c) {
let n = index(startIndex, offsetBy: i)
replaceSubrange(n...n, with: "\(c)")
}
}
}
public extension Substring {
public subscript (i: Int) -> Character {
get {
return self[index(startIndex, offsetBy: i)]
}
set (c) {
let n = index(startIndex, offsetBy: i)
replaceSubrange(n...n, with: "\(c)")
}
}
public subscript (bounds: CountableRange<Int>) -> Substring {
get {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[start ..< end]
}
set (s) {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
replaceSubrange(start ..< end, with: s)
}
}
public subscript (bounds: CountableClosedRange<Int>) -> Substring {
get {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[start ... end]
}
set (s) {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
replaceSubrange(start ... end, with: s)
}
}
public subscript (bounds: CountablePartialRangeFrom<Int>) -> Substring {
get {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(endIndex, offsetBy: -1)
return self[start ... end]
}
set (s) {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(endIndex, offsetBy: -1)
replaceSubrange(start ... end, with: s)
}
}
public subscript (bounds: PartialRangeThrough<Int>) -> Substring {
get {
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[startIndex ... end]
}
set (s) {
let end = index(startIndex, offsetBy: bounds.upperBound)
replaceSubrange(startIndex ..< end, with: s)
}
}
public subscript (bounds: PartialRangeUpTo<Int>) -> Substring {
get {
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[startIndex ..< end]
}
set (s) {
let end = index(startIndex, offsetBy: bounds.upperBound)
replaceSubrange(startIndex ..< end, with: s)
}
}
public subscript (i: Int) -> String {
get {
return "\(self[index(startIndex, offsetBy: i)])"
}
set (c) {
let n = index(startIndex, offsetBy: i)
replaceSubrange(n...n, with: "\(c)")
}
}
public subscript (bounds: CountableRange<Int>) -> String {
get {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
return "\(self[start ..< end])"
}
set (s) {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
replaceSubrange(start ..< end, with: s)
}
}
public subscript (bounds: CountableClosedRange<Int>) -> String {
get {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
return "\(self[start ... end])"
}
set (s) {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
replaceSubrange(start ... end, with: s)
}
}
public subscript (bounds: CountablePartialRangeFrom<Int>) -> String {
get {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(endIndex, offsetBy: -1)
return "\(self[start ... end])"
}
set (s) {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(endIndex, offsetBy: -1)
replaceSubrange(start ... end, with: s)
}
}
public subscript (bounds: PartialRangeThrough<Int>) -> String {
get {
let end = index(startIndex, offsetBy: bounds.upperBound)
return "\(self[startIndex ... end])"
}
set (s) {
let end = index(startIndex, offsetBy: bounds.upperBound)
replaceSubrange(startIndex ... end, with: s)
}
}
public subscript (bounds: PartialRangeUpTo<Int>) -> String {
get {
let end = index(startIndex, offsetBy: bounds.upperBound)
return "\(self[startIndex ..< end])"
}
set (s) {
let end = index(startIndex, offsetBy: bounds.upperBound)
replaceSubrange(startIndex ..< end, with: s)
}
}
public subscript (i: Int) -> Substring {
get {
return Substring("\(self[index(startIndex, offsetBy: i)])")
}
set (c) {
let n = index(startIndex, offsetBy: i)
replaceSubrange(n...n, with: "\(c)")
}
}
}
Solution 31 - Swift
Swift 4.2
This answer is ideal because it extends String and all of its Subsequences (Substring) in one extension
public extension StringProtocol {
public subscript (i: Int) -> Element {
return self[index(startIndex, offsetBy: i)]
}
public subscript (bounds: CountableClosedRange<Int>) -> SubSequence {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[start...end]
}
public subscript (bounds: CountableRange<Int>) -> SubSequence {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[start..<end]
}
public subscript (bounds: PartialRangeUpTo<Int>) -> SubSequence {
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[startIndex..<end]
}
public subscript (bounds: PartialRangeThrough<Int>) -> SubSequence {
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[startIndex...end]
}
public subscript (bounds: CountablePartialRangeFrom<Int>) -> SubSequence {
let start = index(startIndex, offsetBy: bounds.lowerBound)
return self[start..<endIndex]
}
}
Usage
var str = "Hello, playground"
print(str[5...][...5][0])
// Prints ","
Solution 32 - Swift
Swift's String type does not provide a characterAtIndex method because there are several ways a Unicode string could be encoded. Are you going with UTF8, UTF16, or something else?
You can access the CodeUnit collections by retrieving the String.utf8 and String.utf16 properties. You can also access the UnicodeScalar collection by retrieving the String.unicodeScalars property.
In the spirit of NSString's implementation, I'm returning a unichar type.
extension String
{
func characterAtIndex(index:Int) -> unichar
{
return self.utf16[index]
}
// Allows us to use String[index] notation
subscript(index:Int) -> unichar
{
return characterAtIndex(index)
}
}
let text = "Hello Swift!"
let firstChar = text[0]
Solution 33 - Swift
In order to feed the subject and show swift subscript possibilities, here's a little string "substring-toolbox" subscript based
These methods are safe and never go over string indexes
extension String {
// string[i] -> one string char
subscript(pos: Int) -> String { return String(Array(self)[min(self.length-1,max(0,pos))]) }
// string[pos,len] -> substring from pos for len chars on the left
subscript(pos: Int, len: Int) -> String { return self[pos, len, .pos_len, .left2right] }
// string[pos, len, .right2left] -> substring from pos for len chars on the right
subscript(pos: Int, len: Int, way: Way) -> String { return self[pos, len, .pos_len, way] }
// string[range] -> substring form start pos on the left to end pos on the right
subscript(range: Range<Int>) -> String { return self[range.startIndex, range.endIndex, .start_end, .left2right] }
// string[range, .right2left] -> substring start pos on the right to end pos on the left
subscript(range: Range<Int>, way: Way) -> String { return self[range.startIndex, range.endIndex, .start_end, way] }
var length: Int { return countElements(self) }
enum Mode { case pos_len, start_end }
enum Way { case left2right, right2left }
subscript(var val1: Int, var val2: Int, mode: Mode, way: Way) -> String {
if mode == .start_end {
if val1 > val2 { let val=val1 ; val1=val2 ; val2=val }
val2 = val2-val1
}
if way == .left2right {
val1 = min(self.length-1, max(0,val1))
val2 = min(self.length-val1, max(1,val2))
} else {
let val1_ = val1
val1 = min(self.length-1, max(0, self.length-val1_-val2 ))
val2 = max(1, (self.length-1-val1_)-(val1-1) )
}
return self.bridgeToObjectiveC().substringWithRange(NSMakeRange(val1, val2))
//-- Alternative code without bridge --
//var range: Range<Int> = pos...(pos+len-1)
//var start = advance(startIndex, range.startIndex)
//var end = advance(startIndex, range.endIndex)
//return self.substringWithRange(Range(start: start, end: end))
}
}
println("0123456789"[3]) // return "3"
println("0123456789"[3,2]) // return "34"
println("0123456789"[3,2,.right2left]) // return "56"
println("0123456789"[5,10,.pos_len,.left2right]) // return "56789"
println("0123456789"[8,120,.pos_len,.right2left]) // return "01"
println("0123456789"[120,120,.pos_len,.left2right]) // return "9"
println("0123456789"[0...4]) // return "01234"
println("0123456789"[0..4]) // return "0123"
println("0123456789"[0...4,.right2left]) // return "56789"
println("0123456789"[4...0,.right2left]) // return "678" << because ??? range can wear endIndex at 0 ???
Solution 34 - Swift
A python-like solution, which allows you to use negative index,
var str = "Hello world!"
str[-1] // "!"
could be:
extension String {
subscript (var index:Int)->Character{
get {
let n = distance(self.startIndex, self.endIndex)
index %= n
if index < 0 { index += n }
return self[advance(startIndex, index)]
}
}
}
By the way, it may be worth it to transpose the whole python's slice notation
Solution 35 - Swift
You can also convert String to Array of Characters like that:
let text = "My Text"
let index = 2
let charSequence = text.unicodeScalars.map{ Character($0) }
let char = charSequence[index]
This is the way to get char at specified index in constant time.
The example below doesn't run in constant time, but requires linear time. So If You have a lot of searching in String by index use the method above.
let char = text[text.startIndex.advancedBy(index)]
Solution 36 - Swift
I wanted to point out that if you have a large string and need to randomly access many characters from it, you may want to pay the extra memory cost and convert the string to an array for better performance:
// Pay up front for O(N) memory
let chars = Array(veryLargeString.characters)
for i in 0...veryLargeNumber {
// Benefit from O(1) access
print(chars[i])
}
Solution 37 - Swift
In Swift 3 without extensions to the String class, as simple as I can make it!
let myString = "abcedfg"
let characterLocationIndex = myString.index(myString.startIndex, offsetBy: 3)
let myCharacter = myString[characterLocationIndex]
myCharacter is "3" in this example.
Solution 38 - Swift
Using characters would do the job. You can quickly convert the String to an array of characters that can be manipulated by the CharacterView methods.
Example:
let myString = "Hello World!"
let myChars = myString.characters
(full CharacterView doc)
(tested in Swift 3)
Solution 39 - Swift
There's an alternative, explained in String manifesto
extension String : BidirectionalCollection {
subscript(i: Index) -> Character { return characters[i] }
}
Solution 40 - Swift
Get the first letter:
first(str) // retrieve first letter
More here: http://sketchytech.blogspot.com/2014/08/swift-pure-swift-method-for-returning.html
Solution 41 - Swift
You could use SwiftString (https://github.com/amayne/SwiftString) to do this.
"Hello, world!"[0] // H
"Hello, world!"[0...4] // Hello
DISCLAIMER: I wrote this extension
Solution 42 - Swift
Allows Negative Indices
Its always useful not always having to write string[string.length - 1] to get the last character when using a subscript extension. This (Swift 3) extension allows for negative indices, Range and CountableClosedRange.
extension String {
var count: Int { return self.characters.count }
subscript (i: Int) -> Character {
// wraps out of bounds indices
let j = i % self.count
// wraps negative indices
let x = j < 0 ? j + self.count : j
// quick exit for first
guard x != 0 else {
return self.characters.first!
}
// quick exit for last
guard x != count - 1 else {
return self.characters.last!
}
return self[self.index(self.startIndex, offsetBy: x)]
}
subscript (r: Range<Int>) -> String {
let lb = r.lowerBound
let ub = r.upperBound
// quick exit for one character
guard lb != ub else { return String(self[lb]) }
return self[self.index(self.startIndex, offsetBy: lb)..<self.index(self.startIndex, offsetBy: ub)]
}
subscript (r: CountableClosedRange<Int>) -> String {
return self[r.lowerBound..<r.upperBound + 1]
}
}
How you can use it:
var text = "Hello World"
text[-1] // d
text[2] // l
text[12] // e
text[0...4] // Hello
text[0..<4] // Hell
For the more thorough Programmer: Include a guard against empty Strings in this extension
subscript (i: Int) -> Character {
guard self.count != 0 else { return '' }
...
}
subscript (r: Range<Int>) -> String {
guard self.count != 0 else { return "" }
...
}
Solution 43 - Swift
Swift 3:
extension String {
func substring(fromPosition: UInt, toPosition: UInt) -> String? {
guard fromPosition <= toPosition else {
return nil
}
guard toPosition < UInt(characters.count) else {
return nil
}
let start = index(startIndex, offsetBy: String.IndexDistance(fromPosition))
let end = index(startIndex, offsetBy: String.IndexDistance(toPosition) + 1)
let range = start..<end
return substring(with: range)
}
}
"ffaabbcc".substring(fromPosition: 2, toPosition: 5) // return "aabb"
Solution 44 - Swift
Include this extension in your project
extension String{
func trim() -> String
{
return self.trimmingCharacters(in: NSCharacterSet.whitespaces)
}
var length: Int {
return self.count
}
subscript (i: Int) -> String {
return self[i ..< i + 1]
}
func substring(fromIndex: Int) -> String {
return self[min(fromIndex, length) ..< length]
}
func substring(toIndex: Int) -> String {
return self[0 ..< max(0, toIndex)]
}
subscript (r: Range<Int>) -> String {
let range = Range(uncheckedBounds: (lower: max(0, min(length, r.lowerBound)),
upper: min(length, max(0, r.upperBound))))
let start = index(startIndex, offsetBy: range.lowerBound)
let end = index(start, offsetBy: range.upperBound - range.lowerBound)
return String(self[start ..< end])
}
func substring(fromIndex: Int, toIndex:Int)->String{
let startIndex = self.index(self.startIndex, offsetBy: fromIndex)
let endIndex = self.index(startIndex, offsetBy: toIndex-fromIndex)
return String(self[startIndex...endIndex])
}
An then use the function like this
let str = "Sample-String"
let substring = str.substring(fromIndex: 0, toIndex: 0) //returns S
let sampleSubstr = str.substring(fromIndex: 0, toIndex: 5) //returns Sample
Solution 45 - Swift
Check this is Swift 4
let myString = "LOVE"
self.textField1.text = String(Array(myString)[0])
self.textField2.text = String(Array(myString)[1])
self.textField3.text = String(Array(myString)[2])
self.textField4.text = String(Array(myString)[3])
Solution 46 - Swift
Swift 5.1
Here might be the easiest solution out of all these answers.
Add this extension:
extension String {
func retrieveFirstCharacter() -> String? {
guard self.count > 0 else { return nil }
let numberOfCharacters = self.count
return String(self.dropLast(numberOfCharacters - 1))
}
}
Solution 47 - Swift
prob one of the best and simpliest way
let yourString = "thisString"
print(Array(yourString)[8])
puts each letters of your string into arrrays and then you sellect the 9th one