本文介绍了“不完全通用字符名称”与stringWithUTF8String的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

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<$ p> NSString * s = [NSString stringWithUTF8String: \U0627\U0644\U0641\U0631\U0646];
NSLog(@%@,s);

我得到编译错误:

 不完整的通用字符名称

注意,它有时工作正常:

  NSString * UAE = [NSString stringWithUTF8String:\U0627\U0644\U0641\U0631\U0646] ; 
NSLog(@%@,UAE);

和输出:

 الامارات

请帮助。

解决方案

\U和\u不是一回事。 \U转义需要8(十六进制)数字而不是4。



这应该工作:

  NSString * s = [NSString stringWithUTF8String:\\\ا\\\ل\\\ف\\\ر\\\ن]; 


when i try to convert form utf-8 string to NSString like so:

NSString *s = [NSString stringWithUTF8String:"\U0627\U0644\U0641\U0631\U0646"];
NSLog(@"%@", s);

i get the compile error:

incomplete universal character name

note that it sometime just works fine:

NSString *UAE = [NSString stringWithUTF8String:"\U0627\U0644\U0641\U0631\U0646"];
    NSLog(@"%@", UAE);

and the output:

الامارات

so why is that happening? please help.

解决方案

\U and \u are not the same thing. The \U escape expects 8 (hex) digits instead of 4.

This should work:

NSString *s = [NSString stringWithUTF8String:"\u0627\u0644\u0641\u0631\u0646"];

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1403页,肝出来的..

09-08 15:46