如何在Objective-C中将NSString转换为HEX String？ [英] How to convert NSString to HEX String in Objective-C?

问题描述

我有一个NSString，其字符串类似于hello。

现在我想将字符串转换为另一个显示十六进制字符串的NSString对象。怎么做？

解决方案

嗯 - 除了可以在其他地方找到的显而易见的东西之外 - 怎么样：

  NSString * str = @Hello World; 
 
 NSString * hexStr = [NSString stringWithFormat：@％@，
 [NSData dataWithBytes：[str cStringUsingEncoding：NSUTF8StringEncoding] 
 length：strlen（[str cStringUsingEncoding：NSUTF8StringEncoding] ）]]; 
 
 for（NSString * toRemove in [NSArray arrayWithObjects：@<，@>，@，nil]）
 hexStr = [hexStr stringByReplacingOccurrencesOfString：toRemove withString： @ ]; 
 
 NSLog（@％@，hexStr）; 
  

这应该给出类似

$ b <$ p的输出$ p> 48656c6c6f20576f726c64


优化这是留给读者的练习： ）：）

I have a NSString with string like "hello".

Now I want to convert the string into another NSString object which shows a hex string. How to do that ?

解决方案

Hmm - apart from the obvious which can be found elsewhere - how about something like:

    NSString * str = @"Hello World";

    NSString * hexStr = [NSString stringWithFormat:@"%@",
                         [NSData dataWithBytes:[str cStringUsingEncoding:NSUTF8StringEncoding] 
                                        length:strlen([str cStringUsingEncoding:NSUTF8StringEncoding])]];

    for(NSString * toRemove in [NSArray arrayWithObjects:@"<", @">", @" ", nil]) 
        hexStr = [hexStr stringByReplacingOccurrencesOfString:toRemove withString:@""];

    NSLog(@"%@", hexStr);

which should give an output like

    48656c6c6f20576f726c64

Optimising this is left as an exercise to the reader :) :)

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