14

可能重复：
如何将字节数组转换为十六进制字符串，反之亦然？

我需要一种有效且快速的方法来进行这种转换。我尝试了两种不同的方法，但它们对我来说不够有效。对于具有大量数据的应用程序，是否有任何其他快速方法可以实时完成此任务？

  public byte[] StringToByteArray(string hex)
    {
        return Enumerable.Range(0, hex.Length / 2).Select(x => Byte.Parse(hex.Substring(2 * x, 2), NumberStyles.HexNumber)).ToArray(); 
    }

上面的那个对我来说更有效率。

 public static byte[] stringTobyte(string hexString)
    {
        try
        {
            int bytesCount = (hexString.Length) / 2;
            byte[] bytes = new byte[bytesCount];
            for (int x = 0; x < bytesCount; ++x)
            {
                bytes[x] = Convert.ToByte(hexString.Substring(x * 2, 2), 16);
            }
            return bytes;
        }
        catch
        {
            throw;
        }
4

3 回答 3

28

If you really need efficiency then:

  • Don't create substrings
  • Don't create an iterator

Or, and get rid of try blocks which only have a catch block which rethrows... for simplicity rather than efficiency though.

This would be a pretty efficient version:

public static byte[] ParseHex(string hexString)
{
    if ((hexString.Length & 1) != 0)
    {
        throw new ArgumentException("Input must have even number of characters");
    }
    int length = hexString.Length / 2;
    byte[] ret = new byte[length];
    for (int i = 0, j = 0; i < length; i++)
    {
        int high = ParseNybble(hexString[j++]);
        int low = ParseNybble(hexString[j++]);
        ret[i] = (byte) ((high << 4) | low);
    }

    return ret;
}

private static int ParseNybble(char c)
{
    // TODO: Benchmark using if statements instead
    switch (c)
    {
        case '0': case '1': case '2': case '3': case '4':
        case '5': case '6': case '7': case '8': case '9':
            return c - '0';
        case 'a': case 'b': case 'c': case 'd': case 'e': case 'f':
            return c - ('a' - 10);
        case 'A': case 'B': case 'C': case 'D': case 'E': case 'F':
            return c - ('A' - 10);
        default:
            throw new ArgumentException("Invalid nybble: " + c);
    }
    return c;
}

The TODO refers to an alternative like this. I haven't measured which is faster.

private static int ParseNybble(char c)
{
    if (c >= '0' && c <= '9')
    {
        return c - '0';
    }
    c = (char) (c & ~0x20);
    if (c >= 'A' && c <= 'F')
    {
        return c - ('A' - 10);
    }
    throw new ArgumentException("Invalid nybble: " + c);
}
于 2013-01-15T06:58:37.063 回答
5

我从另一个问题中获取了基准代码，并对其进行了重新设计以测试此处给出的十六进制到字节方法：

HexToBytesJon: 36979.7 average ticks (over 150 runs)
HexToBytesJon2: 35886.4 average ticks (over 150 runs)
HexToBytesJonCiC: 31230.2 average ticks (over 150 runs)
HexToBytesJase: 15359.1 average ticks (over 150 runs)

HexToBytesJon是Jon的第一个版本，HexToBytesJon2也是第二个变体。 HexToBytesJonCiC是 Jon 的版本，带有CodesInChaos的建议代码。 HexToBytesJase是我的尝试，基于上述两种方法，但使用替代的 nybble 转换，它避免了错误检查和分支：

    public static byte[] HexToBytesJase(string hexString)
    {
        if ((hexString.Length & 1) != 0)
        {
            throw new ArgumentException("Input must have even number of characters");
        }
        byte[] ret = new byte[hexString.Length/2];
        for (int i = 0; i < ret.Length; i++)
        {
            int high = hexString[i*2];
            int low = hexString[i*2+1];
            high = (high & 0xf) + ((high & 0x40) >> 6) * 9;
            low = (low & 0xf) + ((low & 0x40) >> 6) * 9;

            ret[i] = (byte)((high << 4) | low);
        }

        return ret;
    }
于 2013-01-15T10:19:41.537 回答
5

作为乔恩if基于的变体ParseNybble：

public static byte[] ParseHex(string hexString)
{
    if ((hexString.Length & 1) != 0)
    {
        throw new ArgumentException("Input must have even number of characters");
    }
    byte[] ret = new byte[hexString.Length / 2];
    for (int i = 0; i < ret.Length; i++)
    {
        int high = ParseNybble(hexString[i*2]);
        int low = ParseNybble(hexString[i*2+1]);
        ret[i] = (byte) ((high << 4) | low);
    }

    return ret;
}

private static int ParseNybble(char c)
{
    unchecked
    {
        uint i = (uint)(c - '0');
        if(i < 10)
            return (int)i;
        i = ((uint)c & ~0x20u) - 'A';
        if(i < 6)
            return (int)i+10;
        throw new ArgumentException("Invalid nybble: " + c);
    }
}
于 2013-01-15T09:55:40.197 回答