Convert String.Index to Int or Range<String.Index> to NSRange

Question

So I've found issues relating to the case of converting NSRange to Range<String.Index>, but I've actually run into the opposite problem.

Quite simply, I have a String and a Range<String.Index> and need to convert the latter into an NSRange for use with an older function.

So far my only workaround has been to grab a substring instead like so:

func foo(theString: String, inRange: Range<String.Index>?) -> Bool {
    let theSubString = (nil == inRange) ? theString : theString.substringWithRange(inRange!)
    return olderFunction(theSubString, NSMakeRange(0, countElements(theSubString)))
}

This works of course, but it isn't very pretty, I'd much rather avoid having to grab a sub-string and just use the range itself somehow, is this possible?

Answer 1

let index: Int = string.startIndex.distanceTo(range.startIndex)

Answer 2

I don't know which version introduced it, but in Swift 4.2 you can easily convert between the two.

To convert Range<String.Index> to NSRange:

let range   = s[s.startIndex..<s.endIndex]
let nsRange = NSRange(range, in: s)

To convert NSRange to Range<String.Index>:

let nsRange = NSMakeRange(0, 4)
let range   = Range(nsRange, in: s)

Keep in mind that NSRange is UTF-16 based, while Range<String.Index> is Character based. Hence you can't just use counts and positions to convert between the two!