Swift error when creating a String from a range of a special character

Question

I'm having a problem when I want to create string from a index of string that contains a special character. I'll post you a playground example.

var str = """
circular para poder realizar sus tareas laborales correspondientes a las actividades de comercialización de alimentos
"""

let regex = try? NSRegularExpression(pattern: ".", options: .caseInsensitive)
let results = regex?.matches(in: str, options: .withoutAnchoringBounds, range: NSRange(0..<str.count - 1))
results?.forEach { result in
    let newStr = String(str[Range(result.range, in: str)!])
    print(newStr)
}

Now I get an error when the range of the character "ó" wants to form a string. How can I solve this?

Thanks

The problem is that you must not pass the Swift String character count to NSRange, compare e.g. https://stackoverflow.com/q/27880650/1187415 — Martin R, Sep 11 '20 at 18:28
I don't understand, this code: ''String(text[Range($0.range, in: text)!])" is the same as "String(str[Range(result.range, in: str)!])", son I don't get the solution — Carlos Maria Caraccia, Sep 11 '20 at 18:40
If I use compact map then the ó wont be used because its nil. The range on the regex?.matches is the same, I don't see any changes. — Carlos Maria Caraccia, Sep 11 '20 at 18:47
Even if it worked, what is the purpose/benefit to split a string into characters with Regular Expression? — vadian, Sep 11 '20 at 18:58
No, that’s not the point I want to make it in a pdf, so when I type an expression the characters appear with an annotation, but apparently I cannot capture special chars with regex — Carlos Maria Caraccia, Sep 11 '20 at 19:00
@CarlosMariaCaraccia You need to pass the string utf16 count not the string count when creating your NSRange `NSRange(location: 0, length: str.utf16.count)` — Leo Dabus, Sep 11 '20 at 19:01
@LeoDabus please try it out for yourself with a playground, it does not works, I'm still getting nil, although I compact map it or not. I get nil with the special character. — Carlos Maria Caraccia, Sep 11 '20 at 19:08
@CarlosMariaCaraccia yes but I still can't understand why ns ranges location 102 length 1 and location 103 length 1 return nil when converting to Range — Leo Dabus, Sep 11 '20 at 19:28
It seems that the "." pattern does not work well with *decomposed* Unicode characters. It works if you apply `.precomposedStringWithCanonicalMapping` to the string. – Also the range should be `NSRange(0.. — Martin R, Sep 11 '20 at 19:33
The correct range should be NSRange(location: 102, length: 2) — Leo Dabus, Sep 11 '20 at 19:33
@MartinR `NSRange(str1.startIndex..., in: str1)` won't solve the issue — Leo Dabus, Sep 11 '20 at 19:35
@LeoDabus: As I said, the other problem *seems* to be the decomposed Unicode character. — Martin R, Sep 11 '20 at 19:36
@MartinR let str1 = str.precomposedStringWithCanonicalMapping does not works, although I do NSRange(str1.startIndex..., in: str1), neither — Carlos Maria Caraccia, Sep 11 '20 at 19:39
@LeoDabus I did not get an nsexception, but I still get nil in the character. — Carlos Maria Caraccia, Sep 11 '20 at 19:40
I can only say that it worked in my Xcode 12 (running on macOS 10.15.6) — Martin R, Sep 11 '20 at 19:45
@MartinR I'm running Version 11.5 (11E608c), and it does not works — Carlos Maria Caraccia, Sep 11 '20 at 19:46
You are confusing yourself because you have both `str` and `str1`. — matt, Sep 11 '20 at 20:00
@CarlosMariaCaraccia it did work. it was my mistake. `precomposedStringWithCanonicalMapping ` did fix it. `NSRange(location: 0, length: str1.utf16.count)` worked as well — Leo Dabus, Sep 11 '20 at 20:03
@MartinR Is there any question already that this post can be marked as duplicate otherwise post your comment as an answer. — Leo Dabus, Sep 11 '20 at 20:31
@LeoDabus: I haven't found one yet. I'll am about to write an answer. — Martin R, Sep 11 '20 at 20:32

score 2 · Accepted Answer · answered Sep 11 '20 at 20:43

There are two problems. One is that the NSRange must be created from the count of UTF-16 code units int the string (because that is what NSString uses), compare Swift extract regex matches. So that should be NSRange(0..<str.utf16.count) or NSRange(str.startIndex..., in: str).

The other problem is that the string uses decomposed Unicode characters. Here is a simplified demonstration:

let str = "ó"
print(Array(str.utf16))

let regex = try? NSRegularExpression(pattern: ".", options: .caseInsensitive)
let results = regex?.matches(in: str, range: NSRange(str.startIndex..., in: str))
results?.forEach { result in
    print(result.range, Range(result.range, in: str))
}

// Output:
// [111, 769]
// {0, 1} nil
// {1, 1} nil

NSRegularExpression is an “old” Foundation method and works on an NSString. Here that NSString has two UTF-16 characters, and both are matched by the "." pattern.

The problem is that the returned NSRanges do not correspond to Swift String ranges, and therefore Range(result.range, in: str) returns nil.

A possible solutions is to normalize the string to use only composite Unicode characters:

let str = "ó".precomposedStringWithCanonicalMapping

Now the string has only a single Unicode character, and only a single NSRange is returned. The above test program produces the output

// [243]
// {0, 1} Optional(Range(Swift.String.Index(...)))

Swift error when creating a String from a range of a special character

1 Answers1