How does String substring work in Swift

Question

I've been updating some of my old code and answers with Swift 3 but when I got to Swift Strings and Indexing with substrings things got confusing.

Specifically I was trying the following:

let str = "Hello, playground"
let prefixRange = str.startIndex..<str.startIndex.advancedBy(5)
let prefix = str.substringWithRange(prefixRange)

where the second line was giving me the following error

Value of type 'String' has no member 'substringWithRange'

I see that String does have the following methods now:

str.substring(to: String.Index)
str.substring(from: String.Index)
str.substring(with: Range<String.Index>)

These were really confusing me at first so I started playing around index and range. This is a followup question and answer for substring. I am adding an answer below to show how they are used.

Answer 1

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 them I am choosing to use their API. It helps me to remember that String manipulations are not just simple Int index lookups.

Answer 2

I'm really frustrated at Swift's String access model: everything has to be an Index. All I want is to access the i-th character of the string using Int, not the clumsy index and advancing (which happens to change with every major release). So I made an extension to String:

extension String {
    func index(from: Int) -> Index {
        return self.index(startIndex, offsetBy: from)
    }

    func substring(from: Int) -> String {
        let fromIndex = index(from: from)
        return String(self[fromIndex...])
    }

    func substring(to: Int) -> String {
        let toIndex = index(from: to)
        return String(self[..<toIndex])
    }

    func substring(with r: Range<Int>) -> String {
        let startIndex = index(from: r.lowerBound)
        let endIndex = index(from: r.upperBound)
        return String(self[startIndex..<endIndex])
    }
}

let str = "Hello, playground"
print(str.substring(from: 7))         // playground
print(str.substring(to: 5))           // Hello
print(str.substring(with: 7..<11))    // play

Answer 3

Swift 5 Extension:

extension String {
    subscript(_ range: CountableRange<Int>) -> String {
        let start = index(startIndex, offsetBy: max(0, range.lowerBound))
        let end = index(start, offsetBy: min(self.count - range.lowerBound, 
                                             range.upperBound - range.lowerBound))
        return String(self[start..<end])
    }

    subscript(_ range: CountablePartialRangeFrom<Int>) -> String {
        let start = index(startIndex, offsetBy: max(0, range.lowerBound))
         return String(self[start...])
    }
}

Usage:

let s = "hello"
s[0..<3] // "hel"
s[3...]  // "lo"

Or unicode:

let s = ""
s[0..<1] // ""

Answer 4

Swift 4 & 5:

extension String {
  subscript(_ i: Int) -> String {
    let idx1 = index(startIndex, offsetBy: i)
    let idx2 = index(idx1, offsetBy: 1)
    return String(self[idx1..<idx2])
  }

  subscript (r: Range<Int>) -> String {
    let start = index(startIndex, offsetBy: r.lowerBound)
    let end = index(startIndex, offsetBy: r.upperBound)
    return String(self[start ..< end])
  }

  subscript (r: CountableClosedRange<Int>) -> String {
    let startIndex =  self.index(self.startIndex, offsetBy: r.lowerBound)
    let endIndex = self.index(startIndex, offsetBy: r.upperBound - r.lowerBound)
    return String(self[startIndex...endIndex])
  }
}

How to use it:

"abcde"[0] --> "a"

"abcde"[0...2] --> "abc"

"abcde"[2..<4] --> "cd"

Answer 5

Swift 4

In swift 4 String conforms to Collection. Instead of substring, we should now use a subscript. So if you want to cut out only the word "play" from "Hello, playground", you could do it like this:

var str = "Hello, playground"
let start = str.index(str.startIndex, offsetBy: 7)
let end = str.index(str.endIndex, offsetBy: -6)
let result = str[start..<end] // The result is of type Substring

It is interesting to know, that doing so will give you a Substring instead of a String. This is fast and efficient as Substring shares its storage with the original String. However sharing memory this way can also easily lead to memory leaks.

This is why you should copy the result into a new String, once you want to clean up the original String. You can do this using the normal constructor:

let newString = String(result)

You can find more information on the new Substring class in the [Apple documentation].1

So, if you for example get a Range as the result of an NSRegularExpression, you could use the following extension:

extension String {

    subscript(_ range: NSRange) -> String {
        let start = self.index(self.startIndex, offsetBy: range.lowerBound)
        let end = self.index(self.startIndex, offsetBy: range.upperBound)
        let subString = self[start..<end]
        return String(subString)
    }

}

Answer 6

Here's a function that returns substring of a given substring when start and end indices are provided. For complete reference you can visit the links given below.

func substring(string: String, fromIndex: Int, toIndex: Int) -> String? {
    if fromIndex < toIndex && toIndex < string.count /*use string.characters.count for swift3*/{
        let startIndex = string.index(string.startIndex, offsetBy: fromIndex)
        let endIndex = string.index(string.startIndex, offsetBy: toIndex)
        return String(string[startIndex..<endIndex])
    }else{
        return nil
    }
}

Here's a link to the blog post that I have created to deal with string manipulation in swift. String manipulation in swift (Covers swift 4 as well)

Or you can see this gist on github

Answer 7

I had the same initial reaction. I too was frustrated at how syntax and objects change so drastically in every major release.

However, I realized from experience how I always eventually suffer the consequences of trying to fight "change" like dealing with multi-byte characters which is inevitable if you're looking at a global audience.

So I decided to recognize and respect the efforts exerted by Apple engineers and do my part by understanding their mindset when they came up with this "horrific" approach.

Instead of creating extensions which is just a workaround to make your life easier (I'm not saying they're wrong or expensive), why not figure out how Strings are now designed to work.

For instance, I had this code which was working on Swift 2.2:

let rString = cString.substringToIndex(2)
let gString = (cString.substringFromIndex(2) as NSString).substringToIndex(2)
let bString = (cString.substringFromIndex(4) as NSString).substringToIndex(2)

and after giving up trying to get the same approach working e.g. using Substrings, I finally understood the concept of treating Strings as a bidirectional collection for which I ended up with this version of the same code:

let rString = String(cString.characters.prefix(2))
cString = String(cString.characters.dropFirst(2))
let gString = String(cString.characters.prefix(2))
cString = String(cString.characters.dropFirst(2))
let bString = String(cString.characters.prefix(2))

I hope this contributes...

Answer 8

I'm quite mechanical thinking. Here are the basics...

Swift 4 Swift 5

  let t = "abracadabra"

  let start1 = t.index(t.startIndex, offsetBy:0)
  let   end1 = t.index(t.endIndex, offsetBy:-5)
  let start2 = t.index(t.endIndex, offsetBy:-5)
  let   end2 = t.index(t.endIndex, offsetBy:0)

  let t2 = t[start1 ..< end1]
  let t3 = t[start2 ..< end2]                

  //or a shorter form 

  let t4 = t[..<end1]
  let t5 = t[start2...]

  print("\(t2) \(t3) \(t)")
  print("\(t4) \(t5) \(t)")

  // result:
  // abraca dabra abracadabra

The result is a substring, meaning that it is a part of the original string. To get a full blown separate string just use e.g.

    String(t3)
    String(t4)

This is what I use:

    let mid = t.index(t.endIndex, offsetBy:-5)
    let firstHalf = t[..<mid]
    let secondHalf = t[mid...]

Answer 9

Came across this fairly short and simple way of achieving this.

var str = "Hello, World"
let arrStr = Array(str)
print(arrStr[0..<5]) //["H", "e", "l", "l", "o"]
print(arrStr[7..<12]) //["W", "o", "r", "l", "d"]
print(String(arrStr[0..<5])) //Hello
print(String(arrStr[7..<12])) //World

Answer 10

Same frustration, this should not be that hard...

I compiled this example of getting positions for substring(s) from larger text:

//
// Play with finding substrings returning an array of the non-unique words and positions in text
//
//

import UIKit

let Bigstring = "Why is it so hard to find substrings in Swift3"
let searchStrs : Array<String>? = ["Why", "substrings", "Swift3"]

FindSubString(inputStr: Bigstring, subStrings: searchStrs)


func FindSubString(inputStr : String, subStrings: Array<String>?) ->    Array<(String, Int, Int)> {
    var resultArray : Array<(String, Int, Int)> = []
    for i: Int in 0...(subStrings?.count)!-1 {
        if inputStr.contains((subStrings?[i])!) {
            let range: Range<String.Index> = inputStr.range(of: subStrings![i])!
            let lPos = inputStr.distance(from: inputStr.startIndex, to: range.lowerBound)
            let uPos = inputStr.distance(from: inputStr.startIndex, to: range.upperBound)
            let element = ((subStrings?[i])! as String, lPos, uPos)
            resultArray.append(element)
        }
    }
    for words in resultArray {
        print(words)
    }
    return resultArray
}

returns ("Why", 0, 3) ("substrings", 26, 36) ("Swift3", 40, 46)

Answer 11

I am new in Swift 3, but looking the String (index) syntax for analogy I think that index is like a "pointer" constrained to string and Int can help as an independent object. Using the base + offset syntax , then we can get the i-th character from string with the code bellow:

let s = "abcdefghi"
let i = 2
print (s[s.index(s.startIndex, offsetBy:i)])
// print c

For a range of characters ( indexes) from string using String (range) syntax we can get from i-th to f-th characters with the code bellow:

let f = 6
print (s[s.index(s.startIndex, offsetBy:i )..<s.index(s.startIndex, offsetBy:f+1 )])
//print cdefg

For a substring (range) from a string using String.substring (range) we can get the substring using the code bellow:

print (s.substring (with:s.index(s.startIndex, offsetBy:i )..<s.index(s.startIndex, offsetBy:f+1 ) ) )
//print cdefg

Notes:

1. The i-th and f-th begin with 0.

2. To f-th, I use offsetBY: f + 1, because the range of subscription use ..< (half-open operator), not include the f-th position.

3. Of course must include validate errors like invalid index.

Answer 12

Swift 4+

extension String {
    func take(_ n: Int) -> String {
        guard n >= 0 else {
            fatalError("n should never negative")
        }
        let index = self.index(self.startIndex, offsetBy: min(n, self.count))
        return String(self[..<index])
    }
}

Returns a subsequence of the first n characters, or the entire string if the string is shorter. (inspired by: https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.text/take.html)

Example:

let text = "Hello, World!"
let substring = text.take(5) //Hello

Answer 13

Swift 4

extension String {
    subscript(_ i: Int) -> String {
        let idx1 = index(startIndex, offsetBy: i)
        let idx2 = index(idx1, offsetBy: 1)
        return String(self[idx1..<idx2])
    }
}

let s = "hello"

s[0]    // h
s[1]    // e
s[2]    // l
s[3]    // l
s[4]    // o

Answer 14

I created a simple extension for this (Swift 3)

extension String {
    func substring(location: Int, length: Int) -> String? {
        guard characters.count >= location + length else { return nil }
        let start = index(startIndex, offsetBy: location)
        let end = index(startIndex, offsetBy: location + length)
        return substring(with: start..<end)
    }
}

Answer 15

Heres a more generic implementation:

This technique still uses index to keep with Swift's standards, and imply a full Character.

extension String
{
    func subString <R> (_ range: R) -> String? where R : RangeExpression, String.Index == R.Bound
    {
        return String(self[range])
    }

    func index(at: Int) -> Index
    {
        return self.index(self.startIndex, offsetBy: at)
    }
}

To sub string from the 3rd character:

let item = "Fred looks funny"
item.subString(item.index(at: 2)...) // "ed looks funny"

I've used camel subString to indicate it returns a String and not a Substring.

Answer 16

Building on the above I needed to split a string at a non-printing character dropping the non-printing character. I developed two methods:

var str = "abc\u{1A}12345sdf"
let range1: Range<String.Index> = str.range(of: "\u{1A}")!
let index1: Int = str.distance(from: str.startIndex, to: range1.lowerBound)
let start = str.index(str.startIndex, offsetBy: index1)
let end = str.index(str.endIndex, offsetBy: -0)
let result = str[start..<end] // The result is of type Substring
let firstStr = str[str.startIndex..<range1.lowerBound]

which I put together using some of the answers above.

Because a String is a collection I then did the following:

var fString = String()
for (n,c) in str.enumerated(){

*if c == "\u{1A}" {
    print(fString);
    let lString = str.dropFirst(n + 1)
    print(lString)
    break
   }
 fString += String(c)
}*

Which for me was more intuitive. Which one is best? I have no way of telling They both work with Swift 5

Answer 17

Swift 4

"Substring" (https://developer.apple.com/documentation/swift/substring):

let greeting = "Hi there! It's nice to meet you! "
let endOfSentence = greeting.index(of: "!")!
let firstSentence = greeting[...endOfSentence]
// firstSentence == "Hi there!"

Example of extension String:

private typealias HowDoYouLikeThatElonMusk = String
private extension HowDoYouLikeThatElonMusk {

    subscript(_ from: Character?, _ to: Character?, _ include: Bool) -> String? {
        if let _from: Character = from, let _to: Character = to {
            let dynamicSourceForEnd: String = (_from == _to ? String(self.reversed()) : self)
            guard let startOfSentence: String.Index = self.index(of: _from),
                let endOfSentence: String.Index = dynamicSourceForEnd.index(of: _to) else {
                return nil
            }

            let result: String = String(self[startOfSentence...endOfSentence])
            if include == false {
                guard result.count > 2 else {
                        return nil
                }
                return String(result[result.index(result.startIndex, offsetBy: 1)..<result.index(result.endIndex, offsetBy: -1)])
            }
            return result
        } else if let _from: Character = from {
            guard let startOfSentence: String.Index = self.index(of: _from) else {
                return nil
            }
            let result: String = String(self[startOfSentence...])
            if include == false {
                guard result.count > 1 else {
                    return nil
                }
                return String(result[result.index(result.startIndex, offsetBy: 1)...])
            }
            return result
        } else if let _to: Character = to {
            guard let endOfSentence: String.Index = self.index(of: _to) else {
                    return nil
            }
            let result: String = String(self[...endOfSentence])
            if include == false {
                guard result.count > 1 else {
                    return nil
                }
                return String(result[..<result.index(result.endIndex, offsetBy: -1)])
            }
            return result
        }
        return nil
    }
}

example of using the extension String:

let source =                                   ">>>01234..56789<<<"
// include = true
var from =          source["3", nil, true]  //       "34..56789<<<"
var to =            source[nil, "6", true]  // ">>>01234..56"
var fromTo =        source["3", "6", true]  //       "34..56"
let notFound =      source["a", nil, true]  // nil
// include = false
from =              source["3", nil, false] //        "4..56789<<<"
to =                source[nil, "6", false] // ">>>01234..5"
fromTo =            source["3", "6", false] //        "4..5"
let outOfBounds =   source[".", ".", false] // nil

let str = "Hello, playground"
let hello = str[nil, ",", false] // "Hello"

Answer 18

The specificity of String has mostly been addressed in other answers. To paraphrase: String has a specific Index which is not of type Int because string elements do not have the same size in the general case. Hence, String does not conform to RandomAccessCollection and accessing a specific index implies the traversal of the collection, which is not an O(1) operation.

Many answers have proposed workarounds for using ranges, but they can lead to inefficient code as they use String methods (index(from:), index(:offsetBy:), ...) that are not O(1).

To access string elements like in an array you should use an Array:

let array = Array("Hello, world!")
let letter = array[5]

This is a trade-off, the array creation is an O(n) operation but array accesses are then O(1). You can convert back to a String when you want with String(array).

Answer 19

Swift 5
let desiredIndex: Int = 7 let substring = str[String.Index(encodedOffset: desiredIndex)...]
This substring variable will give you the result.
Simply here Int is converted to Index and then you can split the strings. Unless you will get errors.