如何将字节数组转换为十六进制字符串,反之亦然?


当前回答

用@CodesInChaus补充答案(反向方法)

public static byte[] HexToByteUsingByteManipulation(string s)
{
    byte[] bytes = new byte[s.Length / 2];
    for (int i = 0; i < bytes.Length; i++)
    {
        int hi = s[i*2] - 65;
        hi = hi + 10 + ((hi >> 31) & 7);

        int lo = s[i*2 + 1] - 65;
        lo = lo + 10 + ((lo >> 31) & 7) & 0x0f;

        bytes[i] = (byte) (lo | hi << 4);
    }
    return bytes;
}

说明:

&0x0f还支持小写字母

hi=hi+10+((hi>>31)&7);与以下内容相同:

hi=ch-65+10+((ch-65)>>31)&7);

对于“0”9’与hi=ch-65+10+7相同;其为hi=ch-48(这是因为0xffffff&7)。

对于“A”F’为hi=ch-65+10;(这是因为0x00000000&7)。

对于“a”我们必须使用大数字,所以我们必须通过使用&0x0f使某些位为0,从默认版本中减去32。

65是“A”的代码

48是“0”的代码

7是ASCII表中“9”和“A”之间的字母数(…456789:;<=>?@ABCD…)。

其他回答

另一个快速功能。。。

private static readonly byte[] HexNibble = new byte[] {
    0x0, 0x1, 0x2, 0x3, 0x4, 0x5, 0x6, 0x7,
    0x8, 0x9, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0,
    0x0, 0xA, 0xB, 0xC, 0xD, 0xE, 0xF, 0x0,
    0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0,
    0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0,
    0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0,
    0x0, 0xA, 0xB, 0xC, 0xD, 0xE, 0xF
};

public static byte[] HexStringToByteArray( string str )
{
    int byteCount = str.Length >> 1;
    byte[] result = new byte[byteCount + (str.Length & 1)];
    for( int i = 0; i < byteCount; i++ )
        result[i] = (byte) (HexNibble[str[i << 1] - 48] << 4 | HexNibble[str[(i << 1) + 1] - 48]);
    if( (str.Length & 1) != 0 )
        result[byteCount] = (byte) HexNibble[str[str.Length - 1] - 48];
    return result;
}

我将参加这个比特拨弄比赛,因为我有一个同样使用比特拨弄来解码十六进制的答案。请注意,使用字符数组可能会更快,因为调用StringBuilder方法也需要时间。

public static String ToHex (byte[] data)
{
    int dataLength = data.Length;
    // pre-create the stringbuilder using the length of the data * 2, precisely enough
    StringBuilder sb = new StringBuilder (dataLength * 2);
    for (int i = 0; i < dataLength; i++) {
        int b = data [i];

        // check using calculation over bits to see if first tuple is a letter
        // isLetter is zero if it is a digit, 1 if it is a letter
        int isLetter = (b >> 7) & ((b >> 6) | (b >> 5)) & 1;

        // calculate the code using a multiplication to make up the difference between
        // a digit character and an alphanumerical character
        int code = '0' + ((b >> 4) & 0xF) + isLetter * ('A' - '9' - 1);
        // now append the result, after casting the code point to a character
        sb.Append ((Char)code);

        // do the same with the lower (less significant) tuple
        isLetter = (b >> 3) & ((b >> 2) | (b >> 1)) & 1;
        code = '0' + (b & 0xF) + isLetter * ('A' - '9' - 1);
        sb.Append ((Char)code);
    }
    return sb.ToString ();
}

public static byte[] FromHex (String hex)
{

    // pre-create the array
    int resultLength = hex.Length / 2;
    byte[] result = new byte[resultLength];
    // set validity = 0 (0 = valid, anything else is not valid)
    int validity = 0;
    int c, isLetter, value, validDigitStruct, validDigit, validLetterStruct, validLetter;
    for (int i = 0, hexOffset = 0; i < resultLength; i++, hexOffset += 2) {
        c = hex [hexOffset];

        // check using calculation over bits to see if first char is a letter
        // isLetter is zero if it is a digit, 1 if it is a letter (upper & lowercase)
        isLetter = (c >> 6) & 1;

        // calculate the tuple value using a multiplication to make up the difference between
        // a digit character and an alphanumerical character
        // minus 1 for the fact that the letters are not zero based
        value = ((c & 0xF) + isLetter * (-1 + 10)) << 4;

        // check validity of all the other bits
        validity |= c >> 7; // changed to >>, maybe not OK, use UInt?

        validDigitStruct = (c & 0x30) ^ 0x30;
        validDigit = ((c & 0x8) >> 3) * (c & 0x6);
        validity |= (isLetter ^ 1) * (validDigitStruct | validDigit);

        validLetterStruct = c & 0x18;
        validLetter = (((c - 1) & 0x4) >> 2) * ((c - 1) & 0x2);
        validity |= isLetter * (validLetterStruct | validLetter);

        // do the same with the lower (less significant) tuple
        c = hex [hexOffset + 1];
        isLetter = (c >> 6) & 1;
        value ^= (c & 0xF) + isLetter * (-1 + 10);
        result [i] = (byte)value;

        // check validity of all the other bits
        validity |= c >> 7; // changed to >>, maybe not OK, use UInt?

        validDigitStruct = (c & 0x30) ^ 0x30;
        validDigit = ((c & 0x8) >> 3) * (c & 0x6);
        validity |= (isLetter ^ 1) * (validDigitStruct | validDigit);

        validLetterStruct = c & 0x18;
        validLetter = (((c - 1) & 0x4) >> 2) * ((c - 1) & 0x2);
        validity |= isLetter * (validLetterStruct | validLetter);
    }

    if (validity != 0) {
        throw new ArgumentException ("Hexadecimal encoding incorrect for input " + hex);
    }

    return result;
}

从Java代码转换而来。

有一个简单的一行解决方案尚未提及,它将十六进制字符串转换为字节数组(我们不在乎这里的否定解释,因为这无关紧要):

BigInteger.Parse(str, System.Globalization.NumberStyles.HexNumber).ToByteArray().Reverse().ToArray();

为什么要让它变得复杂?这在Visual Studio 2008中很简单:

C#:

string hex = BitConverter.ToString(YourByteArray).Replace("-", "");

VB:

Dim hex As String = BitConverter.ToString(YourByteArray).Replace("-", "")

另一种方法是使用stackaloc来降低GC内存压力:

static string ByteToHexBitFiddle(byte[] bytes)
{
        var c = stackalloc char[bytes.Length * 2 + 1];
        int b; 
        for (int i = 0; i < bytes.Length; ++i)
        {
            b = bytes[i] >> 4;
            c[i * 2] = (char)(55 + b + (((b - 10) >> 31) & -7));
            b = bytes[i] & 0xF;
            c[i * 2 + 1] = (char)(55 + b + (((b - 10) >> 31) & -7));
        }
        c[bytes.Length * 2 ] = '\0';
        return new string(c);
}