NSRegularExpression does not match anything after CRLF. Why?












0















In Swift 4.2, NSRegularExpression does not match anything after CRLF. Why?



let str = "rnfoo"

let regex = try! NSRegularExpression(pattern: "foo")

print(regex.firstMatch(in: str, range: NSRange(location: 0, length: str.count))) // => nil


If you remove "r" or "n", you get a instance of NSTextCheckingResult.






swift







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  • 3





    It's a range issue. Use NSRange(location: 0, length: str.utf16.count) instead. If you used rnfoooooooooooo (I explicitly add extra char), you would have found it. It's just that your range calculation doesn't cover the "full string range".

    – Larme
    Nov 21 '18 at 10:25


















0















In Swift 4.2, NSRegularExpression does not match anything after CRLF. Why?



let str = "rnfoo"

let regex = try! NSRegularExpression(pattern: "foo")

print(regex.firstMatch(in: str, range: NSRange(location: 0, length: str.count))) // => nil


If you remove "r" or "n", you get a instance of NSTextCheckingResult.






swift







share|improve this question


















  • 3





    It's a range issue. Use NSRange(location: 0, length: str.utf16.count) instead. If you used rnfoooooooooooo (I explicitly add extra char), you would have found it. It's just that your range calculation doesn't cover the "full string range".

    – Larme
    Nov 21 '18 at 10:25
















0












0








0








In Swift 4.2, NSRegularExpression does not match anything after CRLF. Why?



let str = "rnfoo"

let regex = try! NSRegularExpression(pattern: "foo")

print(regex.firstMatch(in: str, range: NSRange(location: 0, length: str.count))) // => nil


If you remove "r" or "n", you get a instance of NSTextCheckingResult.






swift







share|improve this question














In Swift 4.2, NSRegularExpression does not match anything after CRLF. Why?



let str = "rnfoo"

let regex = try! NSRegularExpression(pattern: "foo")

print(regex.firstMatch(in: str, range: NSRange(location: 0, length: str.count))) // => nil


If you remove "r" or "n", you get a instance of NSTextCheckingResult.






swift




swift






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asked Nov 21 '18 at 10:17









YujiYuji

15019




15019








  • 3





    It's a range issue. Use NSRange(location: 0, length: str.utf16.count) instead. If you used rnfoooooooooooo (I explicitly add extra char), you would have found it. It's just that your range calculation doesn't cover the "full string range".

    – Larme
    Nov 21 '18 at 10:25
















  • 3





    It's a range issue. Use NSRange(location: 0, length: str.utf16.count) instead. If you used rnfoooooooooooo (I explicitly add extra char), you would have found it. It's just that your range calculation doesn't cover the "full string range".

    – Larme
    Nov 21 '18 at 10:25










3




3





It's a range issue. Use NSRange(location: 0, length: str.utf16.count) instead. If you used rnfoooooooooooo (I explicitly add extra char), you would have found it. It's just that your range calculation doesn't cover the "full string range".

– Larme
Nov 21 '18 at 10:25







It's a range issue. Use NSRange(location: 0, length: str.utf16.count) instead. If you used rnfoooooooooooo (I explicitly add extra char), you would have found it. It's just that your range calculation doesn't cover the "full string range".

– Larme
Nov 21 '18 at 10:25














1 Answer
1






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oldest

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2














In Swift 4 a new initializer of NSRange was introduced to convert reliably Range<String.Index> to NSRange.



Use it always, it solves your issue.



let str = "rnfoo"

let regex = try! NSRegularExpression(pattern: "foo")

print(regex.firstMatch(in: str, range: NSRange(str.startIndex..., in: str)))





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    In Swift 4 a new initializer of NSRange was introduced to convert reliably Range<String.Index> to NSRange.



    Use it always, it solves your issue.



    let str = "rnfoo"
    
let regex = try! NSRegularExpression(pattern: "foo")
    
print(regex.firstMatch(in: str, range: NSRange(str.startIndex..., in: str)))





    share|improve this answer




























      2














      In Swift 4 a new initializer of NSRange was introduced to convert reliably Range<String.Index> to NSRange.



      Use it always, it solves your issue.



      let str = "rnfoo"
      
let regex = try! NSRegularExpression(pattern: "foo")
      
print(regex.firstMatch(in: str, range: NSRange(str.startIndex..., in: str)))





      share|improve this answer


























        2












        2








        2







        In Swift 4 a new initializer of NSRange was introduced to convert reliably Range<String.Index> to NSRange.



        Use it always, it solves your issue.



        let str = "rnfoo"
        
let regex = try! NSRegularExpression(pattern: "foo")
        
print(regex.firstMatch(in: str, range: NSRange(str.startIndex..., in: str)))





        share|improve this answer













        In Swift 4 a new initializer of NSRange was introduced to convert reliably Range<String.Index> to NSRange.



        Use it always, it solves your issue.



        let str = "rnfoo"
        
let regex = try! NSRegularExpression(pattern: "foo")
        
print(regex.firstMatch(in: str, range: NSRange(str.startIndex..., in: str)))






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 21 '18 at 10:44









        vadianvadian

        148k14159178




        148k14159178
































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