Shortcut to generate an NSRange for entire length of NSString?

Question

Is there a short way to say "entire string" rather than typing out:

NSMakeRange(0, myString.length)]

It seems silly that the longest part of this kind of code is the least important (because I usually want to search/replace within entire string)…

[myString replaceOccurrencesOfString:@"replace_me"
                          withString:replacementString
                             options:NSCaseInsensitiveSearch
                               range:NSMakeRange(0, myString.length)];

If you need more than 5 times, then you can make a method/category, But for under 5 time your `NSMakeRange(0, myString.length)]` will be shorter, isn't it? — Anoop Vaidya, Feb 25 '13 at 08:08

score 47 · Accepted Answer · answered Feb 25 '13 at 08:03

47

Function? Category method?

- (NSRange)fullRange
{
    return (NSRange){0, [self length]};
}

[myString replaceOccurrencesOfString:@"replace_me"
                          withString:replacementString
                             options:NSCaseInsensitiveSearch
                               range:[myString fullRange]];

answered Feb 25 '13 at 08:03

jscs

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9

Hah. Awesome that we both named it `fullRange`, and that we both used the new compound literals, at the same time. – BJ Homer Feb 25 '13 at 08:04
3

Indeed. And commented at the same time! – jscs Feb 25 '13 at 08:05
The literals work just fine, but you could also do return NSMakeRange(0, [self length]); if you wanted. – Brian Sachetta Jun 01 '15 at 15:14

vadian · Answer 2 · 2020-06-16T16:53:46.943

28

Swift 4+, useful for NSRegularExpression and NSAttributedString

extension String {
    var nsRange : NSRange {
        return NSRange(self.startIndex..., in: self)
    }

    func range(from nsRange: NSRange) -> Range<String.Index>? {
        return Range(nsRange, in: self)
    }
}

edited Jun 16 '20 at 16:53

answered May 31 '19 at 08:44

vadian

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score 20 · Answer 3 · answered Feb 25 '13 at 08:03

20

Not that I know of. But you could easily add an NSString category:

@interface NSString (MyRangeExtensions)
- (NSRange)fullRange
@end

@implementation NSString (MyRangeExtensions)
- (NSRange)fullRange {
  return (NSRange){0, self.length};
}

answered Feb 25 '13 at 08:03

BJ Homer

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4

Geez, we even gave it the same name. – jscs Feb 25 '13 at 08:04

score 16 · Answer 4 · edited Jun 20 '20 at 09:12

16

Swift

NSMakeRange(0, str.length)

or as an extension:

extension NSString {
    func fullrange() -> NSRange {
        return NSMakeRange(0, self.length)
    }
}

edited Jun 20 '20 at 09:12

Community

1
1

answered Aug 25 '15 at 18:17

SwiftArchitect

47,376
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3

Correct. But the question is "Shortcut to generate an NSRange for entire length of NSString" – SwiftArchitect Jan 19 '17 at 18:16
3

The question is also about Objective-C. If you use Swift, then also use String. – Tim Vermeulen Aug 03 '17 at 15:16
1

`NSString` may be obsolete in Swift, but `NSAttributedString` isn't, and the same code works for it (aside from the class name, obviously). – John Montgomery Oct 10 '18 at 22:55

score 7 · Answer 5 · edited Jun 20 '20 at 09:12

7

Swift 2:

extension String {
    var fullRange:Range<String.Index> { return startIndex..<endIndex }
}

as in

let swiftRange = "abc".fullRange

or

let nsRange = "abc".fullRange.toRange

edited Jun 20 '20 at 09:12

Community

1
1

answered Jan 01 '16 at 23:41

Chris Conover

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For `"abc".fullRange.toRange`, I get the compile time error: "Value of type 'Range' (aka 'Range') has no member 'toRange'." I'm running Xcode 8.3.3 with Swift 3.1. – ma11hew28 Aug 12 '17 at 15:57

score 6 · Answer 6 · answered Feb 25 '13 at 08:10

6

This is not shorter, but... Oh well

NSRange range = [str rangeOfString:str];

answered Feb 25 '13 at 08:10

Odrakir

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Clever. But I hesitate to introduce what could be a lot of CPU work (matching a long string against itself) for the relatively small benefit of a little less typing for me the programmer. – Basil Bourque Feb 25 '13 at 08:19
And you are right, sometimes a little bit more of code makes it more readable and efficient. – Odrakir Feb 25 '13 at 08:30
I used this as well and ran into a problem: if your string happens to be empty (@""), you get {NSNotFound, 0}. – Shinigami May 08 '14 at 12:52
This function only parse valid ranges as `@"{0, 13}"` etc, it will not create `NSRange` for full length of string, if string does not contain valid range, but rather random string, it returns `{0,0}` range. http://nshipster.com/nsrange/ – Ossir Feb 06 '15 at 12:26

Shortcut to generate an NSRange for entire length of NSString?

6 Answers6

Swift

Swift 2:

Linked