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Can't create a range in Swift 3

I am trying to make a range in Swift 3 that I already had in Swift 2 but it keeps giving me this error: String may not be indexed with 'Int', it has variable size elements

Here is my code:

let range = expireRange!.startIndex.advancedBy(n: 7) ..< expireRange!.startIndex.advancedBy(n: 16)

expiredRange is a Range<Index>?

In Swift 2, I had:

let range = expireRange!.startIndex.advancedBy(7)...expireRange!.startIndex.advancedBy(16)
like image 987
lagoon Avatar asked Jun 28 '16 07:06

lagoon


1 Answers

In Swift 3, "Collections move their index", see A New Model for Collections and Indices on Swift evolution.

Here is an example for String ranges and indices:

let string = "ABCDEFG"
if let range = string.range(of: "CDEF") {
    let lo = string.index(range.lowerBound, offsetBy: 1)
    let hi = string.index(range.lowerBound, offsetBy: 3)
    let subRange = lo ..< hi
    print(string[subRange]) // "DE"
}

The

public func index(_ i: Index, offsetBy n: IndexDistance) -> Index

method is called on the string to calculate the new indices from the range (which has properties lower/upperBound now instead of start/endIndex).

like image 64
Martin R Avatar answered Sep 21 '22 08:09

Martin R