how to convert Range to NSRange in swift 4

Asked Sep 25 '17 at 10:06

Active Sep 25 '17 at 10:09

Viewed 7,946 times

This is my swift 3 code:

extension String {

    func nsRange(from range: Range<Index>) -> NSRange {
         let lower = UTF16View.Index(range.lowerBound, within: utf16)
         let upper = UTF16View.Index(range.upperBound, within: utf16)
         return NSRange(location: utf16.startIndex.distance(to: lower), length: lower!.distance(to: upper))
    }
}

When i migrated to swift 4 it is giving me error distance(to:) is unavailable

Any solution

edited Sep 25 '17 at 10:09

Hamish

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asked Sep 25 '17 at 10:06

chetan panchal

see this https://stackoverflow.com/questions/25138339/nsrange-to-rangestring-index/30404532#30404532 – Anbu.Karthik Sep 25 '17 at 10:07
@JonSnow i want to convert Rang to NSRange – chetan panchal Sep 25 '17 at 10:18
1

@chetanpanchal: This answer https://stackoverflow.com/a/30404532/1187415 to the "duplicate" shows both conversions: NSRange to String Range and vice versa. – Martin R Sep 25 '17 at 10:53
@MartinR Thanks resolved my problem – chetan panchal Sep 26 '17 at 06:04
@Anbu.Karthik Thanks – chetan panchal Sep 26 '17 at 06:04

how to convert Range to NSRange in swift 4

0 Answers0