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PO
 

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    <p>You'll need two character sets:</p> <ul> <li><a href="http://developer.apple.com/library/mac/documentation/Cocoa/Reference/Foundation/Classes/NSCharacterSet_Class/Reference/Reference.html#//apple_ref/occ/clm/NSCharacterSet/letterCharacterSet" rel="nofollow"><code>[NSCharacterSet letterCharacterSet]</code></a></li> <li><a href="http://developer.apple.com/library/mac/documentation/Cocoa/Reference/Foundation/Classes/NSCharacterSet_Class/Reference/Reference.html#//apple_ref/occ/clm/NSCharacterSet/alphanumericCharacterSet" rel="nofollow"><code>[NSCharacterSet alphanumericCharacterSet]</code></a></li> </ul> <p>Using these will enable you to match <em>all</em> letters and numbers in Unicode, not just the English alphabet and digit set. If you really do want only those much-smaller sets, they're easy enough to construct using <a href="http://developer.apple.com/library/mac/documentation/Cocoa/Reference/Foundation/Classes/NSCharacterSet_Class/Reference/Reference.html#//apple_ref/occ/clm/NSCharacterSet/characterSetWithCharactersInString%3a" rel="nofollow"><code>characterSetWithCharactersInString:</code></a> and/or <a href="http://developer.apple.com/library/mac/documentation/Cocoa/Reference/Foundation/Classes/NSCharacterSet_Class/Reference/Reference.html#//apple_ref/occ/clm/NSCharacterSet/characterSetWithRange%3a" rel="nofollow"><code>characterSetWithRange:</code></a>. If you use the latter method, you'll need to make an NSMutableCharacterSet and <a href="http://developer.apple.com/library/mac/documentation/Cocoa/Reference/Foundation/Classes/NSMutableCharacterSet_Class/Reference/Reference.html#//apple_ref/occ/instm/NSMutableCharacterSet/formUnionWithCharacterSet%3a" rel="nofollow">union</a> another character set into it.</p> <p>Once you have your character set, it's easy to <a href="http://developer.apple.com/library/mac/documentation/Cocoa/Reference/Foundation/Classes/NSScanner_Class/Reference/Reference.html#//apple_ref/occ/instm/NSScanner/scanCharactersFromSet%3aintoString%3a" rel="nofollow">scan characters only characters within a set</a> and then, if you want, <a href="http://developer.apple.com/library/mac/documentation/Cocoa/Reference/Foundation/Classes/NSString_Class/Reference/NSString.html#//apple_ref/occ/instm/NSString/stringByAppendingString%3a" rel="nofollow">concatenate one string onto the other</a>.</p>
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    1. POHow to use NSScanner to scan in the format of ^[a-zA-Z_][a-zA-Z0-9_]*
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    1. USPeter Hosey
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    1. POHow to use NSScanner to scan in the format of ^[a-zA-Z_][a-zA-Z0-9_]*
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    1. COThis won't solve my problem. Take, for instance the following string: "this 3 will fail". According to your method there is no way to determine whether "this 3" is equal to "this3". I will chomp on the space without knowing whether I'll be getting an integer, and whether it is separated by a space (or other skipped character).
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      1. USDavid
    2. CO@David: That has nothing to do with the question you asked, which did not include any spaces in the regular expression. Nonetheless, you can do that with NSScanner, too, with different character sets. Don't forget to set the scanner's skip character set to the empty character set if you don't want it to automatically eat spaces.
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      1. USPeter Hosey
    3. COThe point is I want it to function like that regular expression. I do not want it to skip white space, but I also don't want it to read "test 3" if there's a space between the "test" and "3". However, with your explanation, it will do just that.
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