Shortcut to generate an NSRange for entire length of NSString?

问题

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)];

回答1:

Function? Category method?

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

---

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

回答2:

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};
}

回答3:

Swift

NSMakeRange(0, str.length)

or as an extension:

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

回答4:

This is not shorter, but... Oh well

NSRange range = [str rangeOfString:str];

回答5:

Swift 4+, useful for NSRegularExpression and NSAttributedString

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

回答6:

Swift 2:

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

as in

let swiftRange = "abc".fullRange

or

let nsRange = "abc".fullRange.toRange

来源：https://stackoverflow.com/questions/15062458/shortcut-to-generate-an-nsrange-for-entire-length-of-nsstring