How Does String Substring Work in Swift

How does String substring work in Swift

All of the following examples use

var str = "Hello, playground"

Swift 4

Strings got a pretty big overhaul in Swift 4. When you get some substring from a String now, you get a Substring type back rather than a String. Why is this? Strings are value types in Swift. That means if you use one String to make a new one, then it has to be copied over. This is good for stability (no one else is going to change it without your knowledge) but bad for efficiency.

A Substring, on the other hand, is a reference back to the original String from which it came. Here is an image from the documentation illustrating that.

No copying is needed so it is much more efficient to use. However, imagine you got a ten character Substring from a million character String. Because the Substring is referencing the String, the system would have to hold on to the entire String for as long as the Substring is around. Thus, whenever you are done manipulating your Substring, convert it to a String.

let myString = String(mySubstring)

This will copy just the substring over and the memory holding old String can be reclaimed. Substrings (as a type) are meant to be short lived.

Another big improvement in Swift 4 is that Strings are Collections (again). That means that whatever you can do to a Collection, you can do to a String (use subscripts, iterate over the characters, filter, etc).

The following examples show how to get a substring in Swift.

Getting substrings

You can get a substring from a string by using subscripts or a number of other methods (for example, prefix, suffix, split). You still need to use String.Index and not an Int index for the range, though. (See my other answer if you need help with that.)

Beginning of a string

You can use a subscript (note the Swift 4 one-sided range):

let index = str.index(str.startIndex, offsetBy: 5)
let mySubstring = str[..<index] // Hello

or prefix:

let index = str.index(str.startIndex, offsetBy: 5)
let mySubstring = str.prefix(upTo: index) // Hello

or even easier:

let mySubstring = str.prefix(5) // Hello

End of a string

Using subscripts:

let index = str.index(str.endIndex, offsetBy: -10)
let mySubstring = str[index...] // playground

or suffix:

let index = str.index(str.endIndex, offsetBy: -10)
let mySubstring = str.suffix(from: index) // playground

or even easier:

let mySubstring = str.suffix(10) // playground

Note that when using the suffix(from: index) I had to count back from the end by using -10. That is not necessary when just using suffix(x), which just takes the last x characters of a String.

Range in a string

Again we simply use subscripts here.

let start = str.index(str.startIndex, offsetBy: 7)
let end = str.index(str.endIndex, offsetBy: -6)
let range = start..<end

let mySubstring = str[range]  // play

Converting Substring to String

Don't forget, when you are ready to save your substring, you should convert it to a String so that the old string's memory can be cleaned up.

let myString = String(mySubstring)

Using an Int index extension?

I'm hesitant to use an Int based index extension after reading the article Strings in Swift 3 by Airspeed Velocity and Ole Begemann. Although in Swift 4, Strings are collections, the Swift team purposely hasn't used Int indexes. It is still String.Index. This has to do with Swift Characters being composed of varying numbers of Unicode codepoints. The actual index has to be uniquely calculated for every string.

I have to say, I hope the Swift team finds a way to abstract away String.Index in the future. But until then, I am choosing to use their API. It helps me to remember that String manipulations are not just simple Int index lookups.

How can I use String substring in Swift 4? 'substring(to:)' is deprecated: Please use String slicing subscript with a 'partial range from' operator

You should leave one side empty, hence the name "partial range".

let newStr = str[..<index]

The same stands for partial range from operators, just leave the other side empty:

let newStr = str[index...]

Keep in mind that these range operators return a Substring. If you want to convert it to a string, use String's initialization function:

let newStr = String(str[..<index])

You can read more about the new substrings here.

How to get a substring in swift 4?

String.Index is not an integer, and you cannot simply subtract
s.endIndex - 3, as "Collections move their index", see
A New Model for Collections and Indices on Swift evolution.

Care must be taken not to move the index not beyond the valid bounds.
Example:

let s = "aString"

if let upperBound = s.index(s.endIndex, offsetBy: -3, limitedBy: s.startIndex) {
    let subS = String(s[..<upperBound])
} else {
    print("too short")
}

Alternatively,

let upperBound = s.index(s.endIndex, offsetBy: -3, limitedBy: s.startIndex) ?? s.startIndex
let subS = String(s[..<upperBound])

which would print an empty string if s has less then 3 characters.

If you want the initial portion of a string then you can simply do

let subS = String(s.dropLast(3))

or as a mutating method:

var s = "aString"
s.removeLast(min(s.count, 3))
print(s) // "aStr"

Why does change of a substring take no effect on string it subtracted from?

That means that if you change substring or original string they won't be a reference anymore.

https://docs.swift.org/swift-book/LanguageGuide/StringsAndCharacters.html

This performance optimization means you don’t have to pay the
performance cost of copying memory until you modify either the string
or substring

Be careful ususing substring for long time

Storing substrings may, therefore, prolong the lifetime of string data
that is no longer otherwise accessible, which can appear to be memory
leakage.

import UIKit

var str = "Hello, playground"
let index = str.index(str.startIndex, offsetBy: 5)
var mySubstring = str[..<index] // Hello

print(str) // Hello
print(mySubstring) // Hello, playground

str = "Hallo, playground"

print(str) //Hallo, playground
print(mySubstring) //Hello

Swift: How to get substring from start to last index of character

Just accessing backward

The best way is to use substringToIndex combined to the endIndexproperty and the advance global function.

var string1 = "www.stackoverflow.com"

var index1 = advance(string1.endIndex, -4)

var substring1 = string1.substringToIndex(index1)

Looking for a string starting from the back

Use rangeOfString and set options to .BackwardsSearch

var string2 = "www.stackoverflow.com"

var index2 = string2.rangeOfString(".", options: .BackwardsSearch)?.startIndex

var substring2 = string2.substringToIndex(index2!)

No extensions, pure idiomatic Swift

Swift 2.0

advance is now a part of Index and is called advancedBy. You do it like:

var string1 = "www.stackoverflow.com"

var index1 = string1.endIndex.advancedBy(-4)

var substring1 = string1.substringToIndex(index1)

Swift 3.0

You can't call advancedBy on a String because it has variable size elements. You have to use index(_, offsetBy:).

var string1 = "www.stackoverflow.com"

var index1 = string1.index(string1.endIndex, offsetBy: -4)

var substring1 = string1.substring(to: index1)

A lot of things have been renamed. The cases are written in camelCase, startIndex became lowerBound.

var string2 = "www.stackoverflow.com"

var index2 = string2.range(of: ".", options: .backwards)?.lowerBound

var substring2 = string2.substring(to: index2!)

Also, I wouldn't recommend force unwrapping index2. You can use optional binding or map. Personally, I prefer using map:

var substring3 = index2.map(string2.substring(to:))

Swift 4

The Swift 3 version is still valid but now you can now use subscripts with indexes ranges:

let string1 = "www.stackoverflow.com"

let index1 = string1.index(string1.endIndex, offsetBy: -4)

let substring1 = string1[..<index1]

The second approach remains unchanged:

let string2 = "www.stackoverflow.com"

let index2 = string2.range(of: ".", options: .backwards)?.lowerBound

let substring3 = index2.map(string2.substring(to:))

How to get substring with specific ranges in Swift 4?

You can search for substrings using range(of:).

import Foundation

let greeting = "Hello there world!"

if let endIndex = greeting.range(of: "world!")?.lowerBound {
    print(greeting[..<endIndex])
}

outputs:

Hello there 

EDIT:

If you want to separate out the words, there's a quick-and-dirty way and a good way. The quick-and-dirty way:

import Foundation

let greeting = "Hello there world!"

let words = greeting.split(separator: " ")

print(words[1])

And here's the thorough way, which will enumerate all the words in the string no matter how they're separated:

import Foundation

let greeting = "Hello there world!"

var words: [String] = []

greeting.enumerateSubstrings(in: greeting.startIndex..<greeting.endIndex, options: .byWords) { substring, _, _, _ in
    if let substring = substring {
        words.append(substring)
    }
}

print(words[1])

EDIT 2: And if you're just trying to get the 7th through the 11th character, you can do this:

import Foundation

let greeting = "Hello there world!"

let startIndex = greeting.index(greeting.startIndex, offsetBy: 6)
let endIndex = greeting.index(startIndex, offsetBy: 5)

print(greeting[startIndex..<endIndex])

Swift 5: 'substring(to:)' is deprecated

Use String Slicing instead of Substring. like below

let yourString = "dummy string"
yourString[5..<yourString.count] // string

How do you use String.substringWithRange? (or, how do Ranges work in Swift?)

You can use the substringWithRange method. It takes a start and end String.Index.

var str = "Hello, playground"
str.substringWithRange(Range<String.Index>(start: str.startIndex, end: str.endIndex)) //"Hello, playground"

To change the start and end index, use advancedBy(n).

var str = "Hello, playground"
str.substringWithRange(Range<String.Index>(start: str.startIndex.advancedBy(2), end: str.endIndex.advancedBy(-1))) //"llo, playgroun"

You can also still use the NSString method with NSRange, but you have to make sure you are using an NSString like this:

let myNSString = str as NSString
myNSString.substringWithRange(NSRange(location: 0, length: 3))

Note: as JanX2 mentioned, this second method is not safe with unicode strings.

Index of a substring in a string with Swift

edit/update:

Xcode 11.4 • Swift 5.2 or later

import Foundation

extension StringProtocol {
    func index<S: StringProtocol>(of string: S, options: String.CompareOptions = []) -> Index? {
        range(of: string, options: options)?.lowerBound
    }
    func endIndex<S: StringProtocol>(of string: S, options: String.CompareOptions = []) -> Index? {
        range(of: string, options: options)?.upperBound
    }
    func indices<S: StringProtocol>(of string: S, options: String.CompareOptions = []) -> [Index] {
        ranges(of: string, options: options).map(\.lowerBound)
    }
    func ranges<S: StringProtocol>(of string: S, options: String.CompareOptions = []) -> [Range<Index>] {
        var result: [Range<Index>] = []
        var startIndex = self.startIndex
        while startIndex < endIndex,
            let range = self[startIndex...]
                .range(of: string, options: options) {
                result.append(range)
                startIndex = range.lowerBound < range.upperBound ? range.upperBound :
                    index(range.lowerBound, offsetBy: 1, limitedBy: endIndex) ?? endIndex
        }
        return result
    }
}

---

usage:

let str = "abcde"
if let index = str.index(of: "cd") {
    let substring = str[..<index]   // ab
    let string = String(substring)
    print(string)  // "ab\n"
}

---

let str = "Hello, playground, playground, playground"
str.index(of: "play")      // 7
str.endIndex(of: "play")   // 11
str.indices(of: "play")    // [7, 19, 31]
str.ranges(of: "play")     // [{lowerBound 7, upperBound 11}, {lowerBound 19, upperBound 23}, {lowerBound 31, upperBound 35}]

---

case insensitive sample

let query = "Play"
let ranges = str.ranges(of: query, options: .caseInsensitive)
let matches = ranges.map { str[$0] }   //
print(matches)  // ["play", "play", "play"]

---

regular expression sample

let query = "play"
let escapedQuery = NSRegularExpression.escapedPattern(for: query)
let pattern = "\\b\(escapedQuery)\\w+"  // matches any word that starts with "play" prefix

let ranges = str.ranges(of: pattern, options: .regularExpression)
let matches = ranges.map { str[$0] }

print(matches) //  ["playground", "playground", "playground"]

Swift 4: return Substring index from a String

You can use a regex to match your string, assuming that all substrings end with either a ; or the end of string.

This is the regex you should use:

Substring #(\d+): (.+?)(?:;|$)

It captures the substring number into group 1 and the substring into group 2.

You can use it like this:

extension String {
    func substring(withNSRange range: NSRange) -> String {
        return String(string[Range(range, in: self)!])
    }
}

let string = "Substring #1: Hello World!; Substring #2: My name is Tom"
let regex = try! NSRegularExpression(pattern: "Substring #(\\d+): (.+?)(?:;|$)", options: [])
let matches = regex.matches(in: string, options: [], range: NSRange(location: 0, length: string.utf16.count))
let tuples = matches.map { (Int(string.substring(withNSRange: $0.range(at: 1))), string.substring(withNSRange: $0.range(at: 2))) }
let dict = Dictionary(uniqueKeysWithValues: tuples)

// dict will contain something like [1: Hello World!, 2: My name is Tom]

Edit:

Assuming that the custom end of substring is stored in a variable called customStringEnd, you can create the regex like this:

let regex = try! NSRegularExpression(pattern: "Substring #(\\d+): (.+?)(?:\(NSRegularExpression.escapedPattern(for: customStringEnd))|$)", options: [])

---

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