Shortcut to generate an NSRange for entire length of NSString?

ios objective-c nsstring nsrange

35,460

Solution 1

Function? Category method?

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

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

Solution 2

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)
    }
}

Solution 3

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};
}

Solution 4

Swift

NSMakeRange(0, str.length)

or as an extension:

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

Solution 5

Swift 2:

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

as in

let swiftRange = "abc".fullRange

let nsRange = "abc".fullRange.toRange

View more solutions

35,460

Author by

Basil Bourque

SOreadytohelp

Updated on June 17, 2020

Comments

Basil Bourque almost 4 years

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)];

BJ Homer over 11 years

Hah. Awesome that we both named it fullRange, and that we both used the new compound literals, at the same time.
jscs over 11 years

Geez, we even gave it the same name.
jscs over 11 years

Indeed. And commented at the same time!
Basil Bourque over 11 years

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.
Odrakir over 11 years

And you are right, sometimes a little bit more of code makes it more readable and efficient.
Shinigami about 10 years

I used this as well and ran into a problem: if your string happens to be empty (@""), you get {NSNotFound, 0}.
Ossir over 9 years

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. nshipster.com/nsrange
Brian Sachetta almost 9 years

The literals work just fine, but you could also do return NSMakeRange(0, [self length]); if you wanted.
SwiftArchitect over 7 years

Correct. But the question is "Shortcut to generate an NSRange for entire length of NSString"
Tim Vermeulen almost 7 years

The question is also about Objective-C. If you use Swift, then also use String.
ma11hew28 almost 7 years

For "abc".fullRange.toRange, I get the compile time error: "Value of type 'Range<String.Index>' (aka 'Range<String.CharacterView.Index>') has no member 'toRange'." I'm running Xcode 8.3.3 with Swift 3.1.
John Montgomery over 5 years

NSString may be obsolete in Swift, but NSAttributedString isn't, and the same code works for it (aside from the class name, obviously).