How to create range in Swift?

Question

In Objective-c we create range by using NSRange

NSRange range;

So how to create range in Swift?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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)
}

Answer 5

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

Answer 6

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])
        }
    }
}