Swift 3：如何进行字符串范围？

Question

Last night I had to convert my Swift 2.3 code to Swift 3.0 and my code is a mess after the conversion. In Swift 2.3 I had the following code:

let maxChar = 40;
let val = "some long string";
var startRange = val.startIndex;
var endRange = val.startIndex.advancedBy(maxChar, limit: val.endIndex);
let index = val.rangeOfString(" ", options: NSStringCompareOptions.BackwardsSearch , range: startRange...endRange , locale: nil)?.startIndex;

Xcode converted my code to this which doesn't work:

let maxChar = 40;
let val = "some long string";
var startRange = val.startIndex;
var endRange = val.characters.index(val.startIndex, offsetBy: maxChar, limitedBy: val.endIndex);
let index = val.range(of: " ", options: NSString.CompareOptions.backwards , range: startRange...endRange , locale: nil)?.lowerBound

The error is in the parameter range in val.rage , saying No '...' candidates produce the expected contextual result type 'Range?' .

I tried using Range(startRange...endRange) as suggestd in the docs but I'm getting en error saying: connot invoke initiliazer for type ClosedRange<_> with an arguement list of type (ClosedRange) . Seems like I'm missing something fundametnal.

Any help is appreciated. Thanks!

Answer 1

Simple answer: the fundamental thing you are missing is that a closed range is now different from a range. So, change startRange...endRange to startRange..<endRange .

In more detail, here's an abbreviated version of your code (without the maxChar part):

let val = "some long string";
var startRange = val.startIndex;
var endRange = val.endIndex;
let index = val.range(
    of: " ", options: .backwards, range: startRange..<endRange)?.lowerBound
// 9

Now you can use that as a basis to restore your actual desired functionality.

---

However, if all you want to do is split the string, then reinventing the wheel is kind of silly:

let arr = "hey ho ha".characters.split(separator:" ").map{String($0)}
arr // ["hey", "ho", "ha"]