如何将 NSString 转换为十六进制值 [英] How to convert an NSString to hex values

问题描述

我想将常规 NSString 转换为具有(我假设是)ASCII 十六进制值并返回的 NSString.

I'd like to convert a regular NSString into an NSString with the (what I assume are) ASCII hex values and back.

我需要产生与下面的 Java 方法相同的输出，但我似乎无法在 Objective-C 中找到一种方法来做到这一点.我在 C 和 C++ 中找到了一些示例，但我很难将它们融入我的代码中.

I need to produce the same output that the Java methods below do, but I can't seem to find a way to do it in Objective-C. I've found some examples in C and C++ but I've had a hard time working them into my code.

以下是我试图重现的 Java 方法:

Here are the Java methods I'm trying to reproduce:

/**
* Encodes the given string by using the hexadecimal representation of its UTF-8 bytes.
*
* @param s The string to encode.
* @return The encoded string.
*/
public static String utf8HexEncode(String s) {
    if (s == null) {
        return null;
    }
    byte[] utf8;
    try {
        utf8 = s.getBytes(ENCODING_UTF8);
    } catch (UnsupportedEncodingException x) {
        throw new RuntimeException(x);
    }
    return String.valueOf(Hex.encodeHex(utf8));
}

/**
* Decodes the given string by using the hexadecimal representation of its UTF-8 bytes.
*
* @param s The string to decode.
* @return The decoded string.
* @throws Exception If an error occurs.
*/
public static String utf8HexDecode(String s) throws Exception {
if (s == null) {
    return null;
}
    return new String(Hex.decodeHex(s.toCharArray()), ENCODING_UTF8);
} 

更新:感谢drawonward 的回答，这里是我编写的用于创建十六进制NSStrings 的方法.它在 char 声明行上给我一个初始化丢弃来自指针目标类型的限定符"警告，但它有效.

Update: Thanks to drawnonward's answer here's the method I wrote to create the hex NSStrings. It gives me an "Initialization discards qualifiers from pointer target type" warning on the char declaration line, but it works.

- (NSString *)stringToHex:(NSString *)string
{
    char *utf8 = [string UTF8String];
    NSMutableString *hex = [NSMutableString string];
    while ( *utf8 ) [hex appendFormat:@"%02X" , *utf8++ & 0x00FF];

    return [NSString stringWithFormat:@"%@", hex];
}

还没来得及写解码方法.当我这样做时，我会编辑它以将其发布给其他感兴趣的人.

Haven't had time to write the decoding method yet. When I do, I'll edit this to post it for anyone else interested.

Update2: 所以我上面发布的方法实际上并没有输出我正在寻找的东西.它不是以 0-f 格式输出十六进制值，而是输出所有数字.我终于重新开始解决这个问题，并且能够为 NSString 编写一个完全复制我发布的 Java 方法的类别.这是:

Update2: So the method I posted above actually doesn't output what I'm looking for. Instead of outputting hex values in 0-f format, it was instead outputting all numbers. I finally got back to working on this problem and was able to write a category for NSString that exactly duplicates the Java methods I posted. Here it is:

//
//  NSString+hex.h
//  Created by Ben Baron on 10/20/10.
//

@interface NSString (hex) 

    + (NSString *) stringFromHex:(NSString *)str;
    + (NSString *) stringToHex:(NSString *)str;

@end

//
//  NSString+hex.m
//  Created by Ben Baron on 10/20/10.
//

#import "NSString+hex.h"

@implementation NSString (hex)

+ (NSString *) stringFromHex:(NSString *)str 
{   
    NSMutableData *stringData = [[[NSMutableData alloc] init] autorelease];
    unsigned char whole_byte;
    char byte_chars[3] = {'','',''};
    int i;
    for (i=0; i < [str length] / 2; i++) {
        byte_chars[0] = [str characterAtIndex:i*2];
        byte_chars[1] = [str characterAtIndex:i*2+1];
        whole_byte = strtol(byte_chars, NULL, 16);
        [stringData appendBytes:&whole_byte length:1]; 
    }

    return [[[NSString alloc] initWithData:stringData encoding:NSASCIIStringEncoding] autorelease];
}

+ (NSString *) stringToHex:(NSString *)str
{   
    NSUInteger len = [str length];
    unichar *chars = malloc(len * sizeof(unichar));
    [str getCharacters:chars];

    NSMutableString *hexString = [[NSMutableString alloc] init];

    for(NSUInteger i = 0; i < len; i++ )
    {
        [hexString appendString:[NSString stringWithFormat:@"%x", chars[i]]];
    }
    free(chars);

    return [hexString autorelease];
}

@end

推荐答案

对于 Java 的这些行

For these lines of Java

utf8 = s.getBytes(ENCODING_UTF8);
new String(decodedHexString, ENCODING_UTF8);

Objective-C 的等价物是

Objective-C equivalents would be

utf8 = [s UTF8String];
[NSString initWithUTF8String:decodedHexString];

用字符串的十六进制表示生成一个 NSString:

To make an NSString with the hexadecimal representation of a character string:

NSMutableString *hex = [NSMutableString string];
while ( *utf8 ) [hex appendFormat:@"%02X" , *utf8++ & 0x00FF];

您必须制作自己的 decodeHex 函数.只需从字符串中提取两个字符，如果它们有效，则向结果添加一个字节.

You will have to make your own decodeHex function. Just pull two characters out of the string and, if they are valid, add a byte to the result.

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