102

我正在寻找一种很好的可可方法来将 NSData 对象序列化为十六进制字符串。这个想法是在将用于通知的 deviceToken 发送到我的服务器之前对其进行序列化。

我有以下实现，但我认为必须有一些更短更好的方法来做到这一点。

+ (NSString*) serializeDeviceToken:(NSData*) deviceToken
{
    NSMutableString *str = [NSMutableString stringWithCapacity:64];
    int length = [deviceToken length];
    char *bytes = malloc(sizeof(char) * length);

    [deviceToken getBytes:bytes length:length];

    for (int i = 0; i < length; i++)
    {
        [str appendFormat:@"%02.2hhX", bytes[i]];
    }
    free(bytes);

    return str;
}
4

15 回答 15

208

这是我编写的应用于 NSData 的类别。它返回一个表示 NSData 的十六进制 NSString，其中数据可以是任意长度。如果 NSData 为空，则返回一个空字符串。

NSData+Conversion.h

#import <Foundation/Foundation.h>

@interface NSData (NSData_Conversion)

#pragma mark - String Conversion
- (NSString *)hexadecimalString;

@end

NSData+转换.m

#import "NSData+Conversion.h"

@implementation NSData (NSData_Conversion)

#pragma mark - String Conversion
- (NSString *)hexadecimalString {
    /* Returns hexadecimal string of NSData. Empty string if data is empty.   */

    const unsigned char *dataBuffer = (const unsigned char *)[self bytes];

    if (!dataBuffer)
        return [NSString string];

    NSUInteger          dataLength  = [self length];
    NSMutableString     *hexString  = [NSMutableString stringWithCapacity:(dataLength * 2)];

    for (int i = 0; i < dataLength; ++i)
        [hexString appendString:[NSString stringWithFormat:@"%02lx", (unsigned long)dataBuffer[i]]];

    return [NSString stringWithString:hexString];
}

@end

用法：

NSData *someData = ...;
NSString *someDataHexadecimalString = [someData hexadecimalString];

这“可能”比调用[someData description]然后去除空格、< 和 > 更好。剥离字符感觉太“hacky”了。另外，您永远不知道 Apple 将来是否会更改 NSData 的格式-description。

注意：我已经让人们就此答案中的代码许可与我联系。我特此将我在此答案中发布的代码的版权奉献给公共领域。

于 2012-01-31T18:18:46.487 回答
31

这是用于生成十六进制字符串的高度优化的NSData 类别方法。虽然@Dave Gallagher 的回答对于相对较小的尺寸来说已经足够了，但对于大量数据，内存和 CPU 性能会下降。我在 iPhone 5 上使用 2MB 文件对此进行了分析。时间比较为 0.05 秒与 12 秒。这种方法的内存占用可以忽略不计，而另一种方法将堆增加到 70MB！

- (NSString *) hexString
{
    NSUInteger bytesCount = self.length;
    if (bytesCount) {
        const char *hexChars = "0123456789ABCDEF";
        const unsigned char *dataBuffer = self.bytes;
        char *chars = malloc(sizeof(char) * (bytesCount * 2 + 1));       
        if (chars == NULL) {
            // malloc returns null if attempting to allocate more memory than the system can provide. Thanks Cœur
            [NSException raise:NSInternalInconsistencyException format:@"Failed to allocate more memory" arguments:nil];
            return nil;
        }
        char *s = chars;
        for (unsigned i = 0; i < bytesCount; ++i) {
            *s++ = hexChars[((*dataBuffer & 0xF0) >> 4)];
            *s++ = hexChars[(*dataBuffer & 0x0F)];
            dataBuffer++;
        }
        *s = '\0';
        NSString *hexString = [NSString stringWithUTF8String:chars];
        free(chars);
        return hexString;
    }
    return @"";
}
于 2014-08-19T08:05:43.927 回答
17

使用 NSData 的 description 属性不应被认为是对字符串进行 HEX 编码的可接受机制。该属性仅用于描述，可以随时更改。请注意，在 iOS 之前，NSData 描述属性甚至没有以十六进制形式返回它的数据。

很抱歉对解决方案喋喋不休，但重要的是要花精力序列化它，而不需要背负用于数据序列化以外的其他东西的 API。

@implementation NSData (Hex)

- (NSString*)hexString
{
    NSUInteger length = self.length;
    unichar* hexChars = (unichar*)malloc(sizeof(unichar) * (length*2));
    unsigned char* bytes = (unsigned char*)self.bytes;
    for (NSUInteger i = 0; i < length; i++) {
        unichar c = bytes[i] / 16;
        if (c < 10) {
            c += '0';
        } else {
            c += 'A' - 10;
        }
        hexChars[i*2] = c;

        c = bytes[i] % 16;
        if (c < 10) {
            c += '0';
        } else {
            c += 'A' - 10;
        }
        hexChars[i*2+1] = c;
    }
    NSString* retVal = [[NSString alloc] initWithCharactersNoCopy:hexChars length:length*2 freeWhenDone:YES];
    return [retVal autorelease];
}

@end
于 2012-01-25T19:53:10.383 回答
10

这是一种更快的转换方法：

BenchMark（1024 字节数据转换重复 100 次的平均时间）：

Dave Gallagher：~8.070 ms
NSProgrammer：~0.077 ms
Peter：~0.031 ms
This One：~0.017 ms 

@implementation NSData (BytesExtras)

static char _NSData_BytesConversionString_[512] = "000102030405060708090a0b0c0d0e0f101112131415161718191a1b1c1d1e1f202122232425262728292a2b2c2d2e2f303132333435363738393a3b3c3d3e3f404142434445464748494a4b4c4d4e4f505152535455565758595a5b5c5d5e5f606162636465666768696a6b6c6d6e6f707172737475767778797a7b7c7d7e7f808182838485868788898a8b8c8d8e8f909192939495969798999a9b9c9d9e9fa0a1a2a3a4a5a6a7a8a9aaabacadaeafb0b1b2b3b4b5b6b7b8b9babbbcbdbebfc0c1c2c3c4c5c6c7c8c9cacbcccdcecfd0d1d2d3d4d5d6d7d8d9dadbdcdddedfe0e1e2e3e4e5e6e7e8e9eaebecedeeeff0f1f2f3f4f5f6f7f8f9fafbfcfdfeff";

-(NSString*)bytesString
{
    UInt16*  mapping = (UInt16*)_NSData_BytesConversionString_;
    register UInt16 len = self.length;
    char*    hexChars = (char*)malloc( sizeof(char) * (len*2) );

    // --- Coeur's contribution - a safe way to check the allocation
    if (hexChars == NULL) {
    // we directly raise an exception instead of using NSAssert to make sure assertion is not disabled as this is irrecoverable
        [NSException raise:@"NSInternalInconsistencyException" format:@"failed malloc" arguments:nil];
        return nil;
    }
    // ---

    register UInt16* dst = ((UInt16*)hexChars) + len-1;
    register unsigned char* src = (unsigned char*)self.bytes + len-1;

    while (len--) *dst-- = mapping[*src--];

    NSString* retVal = [[NSString alloc] initWithBytesNoCopy:hexChars length:self.length*2 encoding:NSASCIIStringEncoding freeWhenDone:YES];
#if (!__has_feature(objc_arc))
   return [retVal autorelease];
#else
    return retVal;
#endif
}

@end
于 2015-11-03T14:06:31.773 回答
8

函数式 Swift 版本

一个班轮：

let hexString = UnsafeBufferPointer<UInt8>(start: UnsafePointer(data.bytes),
count: data.length).map { String(format: "%02x", $0) }.joinWithSeparator("")

这是一个可重用和自我记录的扩展形式：

extension NSData {
    func base16EncodedString(uppercase uppercase: Bool = false) -> String {
        let buffer = UnsafeBufferPointer<UInt8>(start: UnsafePointer(self.bytes),
                                                count: self.length)
        let hexFormat = uppercase ? "X" : "x"
        let formatString = "%02\(hexFormat)"
        let bytesAsHexStrings = buffer.map {
            String(format: formatString, $0)
        }
        return bytesAsHexStrings.joinWithSeparator("")
    }
}

或者，使用reduce("", combine: +)而不是joinWithSeparator("")被同行视为功能大师。

编辑：我将 String($0, radix: 16) 更改为 String(format: "%02x", $0)，因为一位数字需要填充零

于 2016-03-01T18:02:52.780 回答
7

彼得的答案移植到斯威夫特 

func hexString(data:NSData)->String{
    if data.length > 0 {
        let  hexChars = Array("0123456789abcdef".utf8) as [UInt8];
        let buf = UnsafeBufferPointer<UInt8>(start: UnsafePointer(data.bytes), count: data.length);
        var output = [UInt8](count: data.length*2 + 1, repeatedValue: 0);
        var ix:Int = 0;
        for b in buf {
            let hi  = Int((b & 0xf0) >> 4);
            let low = Int(b & 0x0f);
            output[ix++] = hexChars[ hi];
            output[ix++] = hexChars[low];
        }
        let result = String.fromCString(UnsafePointer(output))!;
        return result;
    }
    return "";
}

迅捷3

func hexString()->String{
    if count > 0 {
        let hexChars = Array("0123456789abcdef".utf8) as [UInt8];
        return withUnsafeBytes({ (bytes:UnsafePointer<UInt8>) -> String in
            let buf = UnsafeBufferPointer<UInt8>(start: bytes, count: self.count);
            var output = [UInt8](repeating: 0, count: self.count*2 + 1);
            var ix:Int = 0;
            for b in buf {
                let hi  = Int((b & 0xf0) >> 4);
                let low = Int(b & 0x0f);
                output[ix] = hexChars[ hi];
                ix += 1;
                output[ix] = hexChars[low];
                ix += 1;
            }
            return String(cString: UnsafePointer(output));
        })
    }
    return "";
}

斯威夫特 5

func hexString()->String{
    if count > 0 {
        let hexChars = Array("0123456789abcdef".utf8) as [UInt8];
        return withUnsafeBytes{ bytes->String in
            var output = [UInt8](repeating: 0, count: bytes.count*2 + 1);
            var ix:Int = 0;
            for b in bytes {
                let hi  = Int((b & 0xf0) >> 4);
                let low = Int(b & 0x0f);
                output[ix] = hexChars[ hi];
                ix += 1;
                output[ix] = hexChars[low];
                ix += 1;
            }
            return String(cString: UnsafePointer(output));
        }
    }
    return "";
}
于 2015-05-23T17:17:37.440 回答
4

我需要解决这个问题，发现这里的答案非常有用，但我担心性能。这些答案中的大多数都涉及从 NSData 中批量复制数据，因此我编写了以下代码以低开销进行转换：

@interface NSData (HexString)
@end

@implementation NSData (HexString)

- (NSString *)hexString {
    NSMutableString *string = [NSMutableString stringWithCapacity:self.length * 3];
    [self enumerateByteRangesUsingBlock:^(const void *bytes, NSRange byteRange, BOOL *stop){
        for (NSUInteger offset = 0; offset < byteRange.length; ++offset) {
            uint8_t byte = ((const uint8_t *)bytes)[offset];
            if (string.length == 0)
                [string appendFormat:@"%02X", byte];
            else
                [string appendFormat:@" %02X", byte];
        }
    }];
    return string;
}

这会为整个结果预先分配字符串中的空间，并避免使用 enumerateByteRangesUsingBlock 复制 NSData 内容。将格式字符串中的 X 更改为 x 将使用小写十六进制数字。如果您不想要字节之间的分隔符，您可以减少语句

if (string.length == 0)
    [string appendFormat:@"%02X", byte];
else
    [string appendFormat:@" %02X", byte];

降到刚刚

[string appendFormat:@"%02X", byte];
于 2014-01-31T20:52:22.147 回答
1

我需要一个适用于可变长度字符串的答案，所以这就是我所做的：

+ (NSString *)stringWithHexFromData:(NSData *)data
{
    NSString *result = [[data description] stringByReplacingOccurrencesOfString:@" " withString:@""];
    result = [result substringWithRange:NSMakeRange(1, [result length] - 2)];
    return result;
}

非常适合作为 NSString 类的扩展。

于 2010-12-02T19:51:26.640 回答
1

您始终可以使用 [yourString uppercaseString] 将数据描述中的字母大写

于 2011-04-29T09:52:47.253 回答
1

将 NSData 序列化/反序列化为 NSString 的更好方法是使用Google Toolbox for Mac Base64 编码器/解码器。只需将包 Foundation 中的 GTMBase64.m、GTMBase64.he GTMDefines.h 文件拖到您的应用项目中，然后执行类似的操作

/**
 * Serialize NSData to Base64 encoded NSString
 */
-(void) serialize:(NSData*)data {

    self.encodedData = [GTMBase64 stringByEncodingData:data];

}

/**
 * Deserialize Base64 NSString to NSData
 */
-(NSData*) deserialize {

    return [GTMBase64 decodeString:self.encodedData];

}
于 2011-06-09T10:31:30.503 回答
1

这是使用 Swift 3 的解决方案

extension Data {

    public var hexadecimalString : String {
        var str = ""
        enumerateBytes { buffer, index, stop in
            for byte in buffer {
                str.append(String(format:"%02x",byte))
            }
        }
        return str
    }

}

extension NSData {

    public var hexadecimalString : String {
        return (self as Data).hexadecimalString
    }

}
于 2016-08-31T11:02:08.453 回答
0

更改%08x以%08X获取大写字符。

于 2010-07-08T15:45:44.497 回答
0 
@implementation NSData (Extn)

- (NSString *)description
{
    NSMutableString *str = [[NSMutableString alloc] init];
    const char *bytes = self.bytes;
    for (int i = 0; i < [self length]; i++) {
        [str appendFormat:@"%02hhX ", bytes[i]];
    }
    return [str autorelease];
}

@end

Now you can call NSLog(@"hex value: %@", data)
于 2013-11-02T01:41:55.767 回答
0

斯威夫特 + 财产。

我更喜欢将十六进制表示作为属性（与bytes和description属性相同）：

extension NSData {

    var hexString: String {

        let buffer = UnsafeBufferPointer<UInt8>(start: UnsafePointer(self.bytes), count: self.length)
        return buffer.map { String(format: "%02x", $0) }.joinWithSeparator("")
    }

    var heXString: String {

        let buffer = UnsafeBufferPointer<UInt8>(start: UnsafePointer(self.bytes), count: self.length)
        return buffer.map { String(format: "%02X", $0) }.joinWithSeparator("")
    }
}

想法是从这个答案中借来的

于 2016-06-30T13:38:12.903 回答
-4 
[deviceToken description]

您需要删除空格。

我个人base64对 进行编码deviceToken，但这是一个品味问题。

于 2009-08-20T12:10:57.463 回答