How to create range in Swift?
SwiftRangeSwift Problem Overview
In Objective-c we create range by using NSRange
NSRange range;
So how to create range in Swift?
Swift Solutions
Solution 1 - Swift
Updated for Swift 4
Swift ranges are more complex than NSRange, and they didn't get any easier in Swift 3. If you want to try to understand the reasoning behind some of this complexity, read this and this. I'll just show you how to create them and when you might use them.
#Closed Ranges: a...b
This range operator creates a Swift range which includes both element a and element b, even if b is the maximum possible value for a type (like Int.max). There are two different types of closed ranges: ClosedRange and CountableClosedRange.
###1. ClosedRange
The elements of all ranges in Swift are comparable (ie, they conform to the Comparable protocol). That allows you to access the elements in the range from a collection. Here is an example:
let myRange: ClosedRange = 1...3
let myArray = ["a", "b", "c", "d", "e"]
myArray[myRange] // ["b", "c", "d"]
However, a ClosedRange is not countable (ie, it does not conform to the Sequence protocol). That means you can't iterate over the elements with a for loop. For that you need the CountableClosedRange.
###2. CountableClosedRange
This is similar to the last one except now the range can also be iterated over.
let myRange: CountableClosedRange = 1...3
let myArray = ["a", "b", "c", "d", "e"]
myArray[myRange] // ["b", "c", "d"]
for index in myRange {
print(myArray[index])
}
#Half-Open Ranges: a..<b
This range operator includes element a but not element b. Like above, there are two different types of half-open ranges: Range and CountableRange.
###1. Range
As with ClosedRange, you can access the elements of a collection with a Range. Example:
let myRange: Range = 1..<3
let myArray = ["a", "b", "c", "d", "e"]
myArray[myRange] // ["b", "c"]
Again, though, you cannot iterate over a Range because it is only comparable, not stridable.
###2. CountableRange
A CountableRange allows iteration.
let myRange: CountableRange = 1..<3
let myArray = ["a", "b", "c", "d", "e"]
myArray[myRange] // ["b", "c"]
for index in myRange {
print(myArray[index])
}
#NSRange
You can (must) still use NSRange at times in Swift (when making attributed strings, for example), so it is helpful to know how to make one.
let myNSRange = NSRange(location: 3, length: 2)
Note that this is location and length, not start index and end index. The example here is similar in meaning to the Swift range 3..<5. However, since the types are different, they are not interchangeable.
#Ranges with Strings
The ... and ..< range operators are a shorthand way of creating ranges. For example:
let myRange = 1..<3
The long hand way to create the same range would be
let myRange = CountableRange<Int>(uncheckedBounds: (lower: 1, upper: 3)) // 1..<3
You can see that the index type here is Int. That doesn't work for String, though, because Strings are made of Characters and not all characters are the same size. (Read this for more info.) An emoji like , for example, takes more space than the letter "b".
Problem with NSRange
Try experimenting with NSRange and an NSString with emoji and you'll see what I mean. Headache.
let myNSRange = NSRange(location: 1, length: 3)
let myNSString: NSString = "abcde"
myNSString.substring(with: myNSRange) // "bcd"
let myNSString2: NSString = "a😀cde"
myNSString2.substring(with: myNSRange) // "😀c" Where is the "d"!?
The smiley face takes two UTF-16 code units to store, so it gives the unexpected result of not including the "d".
Swift Solution
Because of this, with Swift Strings you use Range<String.Index>, not Range<Int>. The String Index is calculated based on a particular string so that it knows if there are any emoji or extended grapheme clusters.
Example
var myString = "abcde"
let start = myString.index(myString.startIndex, offsetBy: 1)
let end = myString.index(myString.startIndex, offsetBy: 4)
let myRange = start..<end
myString[myRange] // "bcd"
myString = "a😀cde"
let start2 = myString.index(myString.startIndex, offsetBy: 1)
let end2 = myString.index(myString.startIndex, offsetBy: 4)
let myRange2 = start2..<end2
myString[myRange2] // "😀cd"
#One-sided Ranges: a... and ...b and ..<b
In Swift 4 things were simplified a bit. Whenever the starting or ending point of a range can be inferred, you can leave it off.
Int
You can use one-sided integer ranges to iterate over collections. Here are some examples from the documentation.
// iterate from index 2 to the end of the array
for name in names[2...] {
print(name)
}
// iterate from the beginning of the array to index 2
for name in names[...2] {
print(name)
}
// iterate from the beginning of the array up to but not including index 2
for name in names[..<2] {
print(name)
}
// the range from negative infinity to 5. You can't iterate forward
// over this because the starting point in unknown.
let range = ...5
range.contains(7) // false
range.contains(4) // true
range.contains(-1) // true
// You can iterate over this but it will be an infinate loop
// so you have to break out at some point.
let range = 5...
String
This also works with String ranges. If you are making a range with str.startIndex or str.endIndex at one end, you can leave it off. The compiler will infer it.
Given
var str = "Hello, playground"
let index = str.index(str.startIndex, offsetBy: 5)
let myRange = ..<index // Hello
You can go from the index to str.endIndex by using ...
var str = "Hello, playground"
let index = str.index(str.endIndex, offsetBy: -10)
let myRange = index... // playground
#See also:
Notes
- You can't use a range you created with one string on a different string.
- As you can see, String ranges are a pain in Swift, but they do make it possibly to deal better with emoji and other Unicode scalars.
#Further Study
Solution 2 - Swift
Xcode 8 beta 2 • Swift 3
let myString = "Hello World"
let myRange = myString.startIndex..<myString.index(myString.startIndex, offsetBy: 5)
let mySubString = myString.substring(with: myRange) // Hello
Xcode 7 • Swift 2.0
let myString = "Hello World"
let myRange = Range<String.Index>(start: myString.startIndex, end: myString.startIndex.advancedBy(5))
let mySubString = myString.substringWithRange(myRange) // Hello
or simply
let myString = "Hello World"
let myRange = myString.startIndex..<myString.startIndex.advancedBy(5)
let mySubString = myString.substringWithRange(myRange) // Hello
Solution 3 - Swift
I find it surprising that, even in Swift 4, there's still no simple native way to express a String range using Int. The only String methods that let you supply an Int as a way of obtaining a substring by range are prefix and suffix.
It is useful to have on hand some conversion utilities, so that we can talk like NSRange when speaking to a String. Here's a utility that takes a location and length, just like NSRange, and returns a Range<String.Index>:
func range(_ start:Int, _ length:Int) -> Range<String.Index> {
let i = self.index(start >= 0 ? self.startIndex : self.endIndex,
offsetBy: start)
let j = self.index(i, offsetBy: length)
return i..<j
}
For example, "hello".range(0,1)" is the Range<String.Index> embracing the first character of "hello". As a bonus, I've allowed negative locations: "hello".range(-1,1)" is the Range<String.Index> embracing the last character of "hello".
It is useful also to convert a Range<String.Index> to an NSRange, for those moments when you have to talk to Cocoa (for example, in dealing with NSAttributedString attribute ranges). Swift 4 provides a native way to do that:
let nsrange = NSRange(range, in:s) // where s is the string
We can thus write another utility where we go directly from a String location and length to an NSRange:
extension String {
func nsRange(_ start:Int, _ length:Int) -> NSRange {
return NSRange(self.range(start,length), in:self)
}
}
Solution 4 - Swift
Use like this
var start = str.startIndex // Start at the string's start index
var end = advance(str.startIndex, 5) // Take start index and advance 5 characters forward
var range: Range<String.Index> = Range<String.Index>(start: start,end: end)
let firstFiveDigit = str.substringWithRange(range)
print(firstFiveDigit)
Output : Hello
Solution 5 - Swift
> (1..<10)
returns...
> Range
Solution 6 - Swift
If anyone want to create NSRange object can create as:
let range: NSRange = NSRange.init(location: 0, length: 5)
this will create range with position 0 and length 5
Solution 7 - Swift
You can use like this
let nsRange = NSRange(location: someInt, length: someInt)
as in
let myNSString = bigTOTPCode as NSString //12345678
let firstDigit = myNSString.substringWithRange(NSRange(location: 0, length: 1)) //1
let secondDigit = myNSString.substringWithRange(NSRange(location: 1, length: 1)) //2
let thirdDigit = myNSString.substringWithRange(NSRange(location: 2, length: 4)) //3456
Solution 8 - Swift
I created the following extension:
extension String {
func substring(from from:Int, to:Int) -> String? {
if from<to && from>=0 && to<self.characters.count {
let rng = self.startIndex.advancedBy(from)..<self.startIndex.advancedBy(to)
return self.substringWithRange(rng)
} else {
return nil
}
}
}
example of use:
print("abcde".substring(from: 1, to: 10)) //nil
print("abcde".substring(from: 2, to: 4)) //Optional("cd")
print("abcde".substring(from: 1, to: 0)) //nil
print("abcde".substring(from: 1, to: 1)) //nil
print("abcde".substring(from: -1, to: 1)) //nil
Solution 9 - Swift
func replace(input: String, start: Int,lenght: Int, newChar: Character) -> String {
var chars = Array(input.characters)
for i in start...lenght {
guard i < input.characters.count else{
break
}
chars[i] = newChar
}
return String(chars)
}
Solution 10 - Swift
I want to do this:
print("Hello"[1...3])
// out: Error
But unfortunately, I can't write a subscript of my own because the loathed one takes up the name space.
We can do this however:
print("Hello"[range: 1...3])
// out: ell
Just add this to your project:
extension String {
subscript(range: ClosedRange<Int>) -> String {
get {
let start = String.Index(utf16Offset: range.lowerBound, in: self)
let end = String.Index(utf16Offset: range.upperBound, in: self)
return String(self[start...end])
}
}
}