Distance(From:To:)' Is Unavailable: Any String View Index Conversion Can Fail in Swift 4; Please Unwrap the Optional Indices

Swift 4: 'distance(to:)' is unavailable

Slicing strings has changed quite a bit in Swift 4. NSRange is generally not the right tool anymore, so how to fix this depends on the rest of the surrounding code. That said, the equivalent code in Swift 4 is this:

NSRange(location:myString.distance(from: myString.startIndex, to: myIndex),length:myLength)

See the String Manifesto for how Strings are intended to work going forward. See SE-0163 and SE-0180 for more of the specific changes in 4.0.

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.

NSRange from Swift Range?

Swift String ranges and NSString ranges are not "compatible".
For example, an emoji like counts as one Swift character, but as two NSString
characters (a so-called UTF-16 surrogate pair).

Therefore your suggested solution will produce unexpected results if the string
contains such characters. Example:

let text = "Long paragraph saying!"
let textRange = text.startIndex..<text.endIndex
let attributedString = NSMutableAttributedString(string: text)

text.enumerateSubstringsInRange(textRange, options: NSStringEnumerationOptions.ByWords, { (substring, substringRange, enclosingRange, stop) -> () in
    let start = distance(text.startIndex, substringRange.startIndex)
    let length = distance(substringRange.startIndex, substringRange.endIndex)
    let range = NSMakeRange(start, length)

    if (substring == "saying") {
        attributedString.addAttribute(NSForegroundColorAttributeName, value: NSColor.redColor(), range: range)
    }
})
println(attributedString)

Output:


Long paragra{
}ph say{
    NSColor = "NSCalibratedRGBColorSpace 1 0 0 1";
}ing!{
}

As you see, "ph say" has been marked with the attribute, not "saying".

Since NS(Mutable)AttributedString ultimately requires an NSString and an NSRange, it is actually
better to convert the given string to NSString first. Then the substringRange
is an NSRange and you don't have to convert the ranges anymore:

let text = "Long paragraph saying!"
let nsText = text as NSString
let textRange = NSMakeRange(0, nsText.length)
let attributedString = NSMutableAttributedString(string: nsText)

nsText.enumerateSubstringsInRange(textRange, options: NSStringEnumerationOptions.ByWords, { (substring, substringRange, enclosingRange, stop) -> () in

    if (substring == "saying") {
        attributedString.addAttribute(NSForegroundColorAttributeName, value: NSColor.redColor(), range: substringRange)
    }
})
println(attributedString)

Output:


Long paragraph {
}saying{
    NSColor = "NSCalibratedRGBColorSpace 1 0 0 1";
}!{
}

Update for Swift 2:

let text = "Long paragraph saying!"
let nsText = text as NSString
let textRange = NSMakeRange(0, nsText.length)
let attributedString = NSMutableAttributedString(string: text)

nsText.enumerateSubstringsInRange(textRange, options: .ByWords, usingBlock: {
    (substring, substringRange, _, _) in

    if (substring == "saying") {
        attributedString.addAttribute(NSForegroundColorAttributeName, value: NSColor.redColor(), range: substringRange)
    }
})
print(attributedString)

Update for Swift 3:

let text = "Long paragraph saying!"
let nsText = text as NSString
let textRange = NSMakeRange(0, nsText.length)
let attributedString = NSMutableAttributedString(string: text)

nsText.enumerateSubstrings(in: textRange, options: .byWords, using: {
    (substring, substringRange, _, _) in

    if (substring == "saying") {
        attributedString.addAttribute(NSForegroundColorAttributeName, value: NSColor.red, range: substringRange)
    }
})
print(attributedString)

Update for Swift 4:

As of Swift 4 (Xcode 9), the Swift standard library
provides method to convert between Range<String.Index> and NSRange.
Converting to NSString is no longer necessary:

let text = "Long paragraph saying!"
let attributedString = NSMutableAttributedString(string: text)

text.enumerateSubstrings(in: text.startIndex..<text.endIndex, options: .byWords) {
    (substring, substringRange, _, _) in
    if substring == "saying" {
        attributedString.addAttribute(.foregroundColor, value: NSColor.red,
                                      range: NSRange(substringRange, in: text))
    }
}
print(attributedString)

Here substringRange is a Range<String.Index>, and that is converted to the
corresponding NSRange with

NSRange(substringRange, in: text)

How to enumerate an enum with String type?

Swift 4.2+

Starting with Swift 4.2 (with Xcode 10), just add protocol conformance to CaseIterable to benefit from allCases. To add this protocol conformance, you simply need to write somewhere:

extension Suit: CaseIterable {}

If the enum is your own, you may specify the conformance directly in the declaration:

enum Suit: String, CaseIterable { case spades = "♠"; case hearts = "♥"; case diamonds = "♦"; case clubs = "♣" }

Then the following code will print all possible values:

Suit.allCases.forEach {
    print($0.rawValue)
}

Compatibility with earlier Swift versions (3.x and 4.x)

If you need to support Swift 3.x or 4.0, you may mimic the Swift 4.2 implementation by adding the following code:

#if !swift(>=4.2)
public protocol CaseIterable {
    associatedtype AllCases: Collection where AllCases.Element == Self
    static var allCases: AllCases { get }
}
extension CaseIterable where Self: Hashable {
    static var allCases: [Self] {
        return [Self](AnySequence { () -> AnyIterator<Self> in
            var raw = 0
            var first: Self?
            return AnyIterator {
                let current = withUnsafeBytes(of: &raw) { $0.load(as: Self.self) }
                if raw == 0 {
                    first = current
                } else if current == first {
                    return nil
                }
                raw += 1
                return current
            }
        })
    }
}
#endif

How do I catch Index out of range in Swift?

As suggested in comments and other answers it is better to avoid this kind of situations. However, in some cases you might want to check if an item exists in an array and if it does safely return it. For this you can use the below Array extension for safely returning an array item.

Swift 5

extension Collection where Indices.Iterator.Element == Index {
    subscript (safe index: Index) -> Iterator.Element? {
        return indices.contains(index) ? self[index] : nil
    }
}

Swift 4

extension Collection where Indices.Iterator.Element == Index {
    subscript (safe index: Index) -> Iterator.Element? {
        return indices.contains(index) ? self[index] : nil
    }
}

Swift 3

extension Collection where Indices.Iterator.Element == Index {
    subscript (safe index: Index) -> Generator.Element? {
        return indices.contains(index) ? self[index] : nil
    }
}

Swift 2

extension Array {
    subscript (safe index: Int) -> Element? {
        return indices ~= index ? self[index] : nil
    }
}

This way you'll never hit Index out of range
You'll have to check if the item is nil

refer this question for more

Trying the Swift3 code in a Playground in Xcode 8.3.2 still leads to a
"crash" when I do let ar = [1,3,4], then let v = ar[5]. Why? – Thomas
Tempelmann May 17 at 17:40

You have to use our customized subscript so instead of let v = ar[5], it wll be let v = ar[safe: 5].

Default getting value from array.

let boo = foo[index]

Add use the customized subscript.

let boo = fee[safe: index]

// And we can warp the result using guard to keep the code going without throwing the exception.
guard let boo = foo[safe: index] else {
  return
}