Question or problem with Swift language programming:
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
How to solve the problem:
Solution 1:
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) -> SubSequence {
let startIndex = index(self.startIndex, offsetBy: range.lowerBound)
return self[startIndex..<index(startIndex, offsetBy: range.count)]
}
subscript(range: ClosedRange) -> SubSequence {
let startIndex = index(self.startIndex, offsetBy: range.lowerBound)
return self[startIndex..<index(startIndex, offsetBy: range.count)]
}
subscript(range: PartialRangeFrom) -> SubSequence { self[index(startIndex, offsetBy: range.lowerBound)...] }
subscript(range: PartialRangeThrough) -> SubSequence { self[...index(startIndex, offsetBy: range.upperBound)] }
subscript(range: PartialRangeUpTo) -> 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 StringProtocol
without success, because the index method does not exist there. 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 implementation
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 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) -> 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:
I just came up with this neat workaround
var firstChar = Array(string)[0]
Solution 4:
Xcode 11 • 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) -> SubSequence { prefix(range.lowerBound+range.count).suffix(range.count) }
subscript(_ range: ClosedRange) -> SubSequence { prefix(range.lowerBound+range.count).suffix(range.count) }
subscript(_ range: PartialRangeThrough) -> SubSequence { prefix(range.upperBound.advanced(by: 1)) }
subscript(_ range: PartialRangeUpTo) -> SubSequence { prefix(range.upperBound) }
subscript(_ range: PartialRangeFrom) -> 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 🇺🇸!!! Hello Brazil 🇧🇷!!!" test[safe: 10] // "🇺🇸" test[11] // "!" test[10...] // "🇺🇸!!! Hello Brazil 🇧🇷!!!" test[10..<12] // "🇺🇸!" test[10...12] // "🇺🇸!!" test[...10] // "Hello USA 🇺🇸" 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 // "🇺🇸!!! Hello Brazil 🇧🇷!!!"
Solution 5:
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]
