Using Range<Index> with NSRange in the NSAttributedString API

Question

I'm attempting to determine the indexes of occurrences of a given string in a String, then generate an NSRange using those indexes in order to add attributes to an NSMutableAttributedString. The problem is rangeOfString returns Range<Index> but addAttributes:range: expects an NSRange. My attempts to create an NSRange from the start and end indexes of the Range have failed, because String.CharacterView.Index is not an Int, thus it will not compile.

How can one use Range<Index> values to create an NSRange?

var originalString = "Hello {world} and those who inhabit it."

let firstBraceIndex = originalString.rangeOfString("{") //Range<Index>
let firstClosingBraceIndex = originalString.rangeOfString("}")

let range = NSMakeRange(firstBraceIndex.startIndex, firstClosingBraceIndex.endIndex)
//compile time error: cannot convert value of type Index to expected argument type Int

let attributedString = NSMutableAttributedString(string: originalString)
attributedString.addAttributes([NSFontAttributeName: boldFont], range: range)

I don't see you calling rangeOfString anywhere. Your code makes no sense. What the heck is originalString("{")???? This cannot possibly be your real code; it is nonsense. — matt
– matt, Commented Oct 4, 2015 at 17:59
My answer remains the same. Where you now have let firstBraceIndex = originalString.rangeOfString("{"), say let firstBraceIndex = (originalString as NSString).rangeOfString("{"). Now you have an NSRange and can carry on from there. — matt
– matt, Commented Oct 4, 2015 at 18:14

matt · Accepted Answer · 2015-10-04 18:03:57Z

8

If you start with your original string cast as a Cocoa NSString:

var originalString = "Hello {world} and those who inhabit it." as NSString

... then your range results will be NSRange and you'll be able to hand them back to Cocoa.

answered Oct 4, 2015 at 18:03

matt

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4 Comments

matt Over a year ago

And see my book, which goes into this topic in some detail: apeth.com/swiftBook/ch03.html#_string

Jordan H Over a year ago

No way to do this with a native Swift String?

matt Over a year ago

This is the way to do it with a native Swift string. I don't see what you're after. I've solved the problem for you. Do you need me to rewrite your whole code and prove it? Just try what I said.

The Cappy Over a year ago

It might be the way to do it, but involves bridging. It doesn't quite use String, and it definitely doesn't use Swift Ranges. It worked for me, but there's more than a whiff of "workaround" to it. It's perfectly clear what bothers him about the solution; there's no need to be abrasive about it.

Marcus · Accepted Answer · 2017-01-03 16:36:39Z

To make an NSRange you need to get the starting location and length of the range as ints. You can do this using the distance(from:to:) method on your originalString:

let rangeStartIndex = firstBraceIndex!.lowerBound
let rangeEndIndex = firstClosingBraceIndex!.upperBound

let start = originalString.distance(from: originalString.startIndex, to: rangeStartIndex)
let length = originalString.distance(from: rangeStartIndex, to: rangeEndIndex)

let nsRange = NSMakeRange(start, length)

let attributedString = NSMutableAttributedString(string: originalString)
attributedString.addAttributes([NSFontAttributeName: boldFont], range: nsRange)

To get the startingLocation, get the distance from the originalString's startIndex, to the starting index of the range you want (which in your case would be the firstBraceIndex.lowerBound if you want to include the {, or firstBraceIndex.upperBound if you don't). Then to get the length of your range, get the distance from your range's starting index to its ending index.

I just force unwrapped your firstBraceIndex and firstClosingBraceIndex to make the code easier to read, but of course it would be better to properly deal with these potentially being nil.

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