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

    // a safe version of the lookup solution:       

    public static string ByteArrayToHexViaLookup32Safe(byte[] bytes, bool withZeroX)
    {
        if (bytes.Length == 0)
        {
            return withZeroX ? "0x" : "";
        }

        int length = bytes.Length * 2 + (withZeroX ? 2 : 0);
        StateSmall stateToPass = new StateSmall(bytes, withZeroX);
        return string.Create(length, stateToPass, (chars, state) =>
        {
            int offset0x = 0;
            if (state.WithZeroX)
            {
                chars[0] = '0';
                chars[1] = 'x';
                offset0x += 2;
            }

            Span<uint> charsAsInts = MemoryMarshal.Cast<char, uint>(chars.Slice(offset0x));
            int targetLength = state.Bytes.Length;
            for (int i = 0; i < targetLength; i += 1)
            {
                uint val = Lookup32[state.Bytes[i]];
                charsAsInts[i] = val;
            }
        });
    }

    private struct StateSmall
    {
        public StateSmall(byte[] bytes, bool withZeroX)
        {
            Bytes = bytes;
            WithZeroX = withZeroX;
        }

        public byte[] Bytes;
        public bool WithZeroX;
    }

其他回答

扩展方法(免责声明:完全未经测试的代码,BTW…):

public static class ByteExtensions
{
    public static string ToHexString(this byte[] ba)
    {
        StringBuilder hex = new StringBuilder(ba.Length * 2);

        foreach (byte b in ba)
        {
            hex.AppendFormat("{0:x2}", b);
        }
        return hex.ToString();
    }
}

使用Tomalak的三种解决方案之一(最后一种是字符串上的扩展方法)。

如果您希望比BitConverter更灵活,但不希望使用那些笨重的90年代风格的显式循环,那么您可以这样做:

String.Join(String.Empty, Array.ConvertAll(bytes, x => x.ToString("X2")));

或者,如果您使用的是.NET 4.0:

String.Concat(Array.ConvertAll(bytes, x => x.ToString("X2")));

(后者来自对原帖子的评论。)

多样性的另一种变化:

public static byte[] FromHexString(string src)
{
    if (String.IsNullOrEmpty(src))
        return null;

    int index = src.Length;
    int sz = index / 2;
    if (sz <= 0)
        return null;

    byte[] rc = new byte[sz];

    while (--sz >= 0)
    {
        char lo = src[--index];
        char hi = src[--index];

        rc[sz] = (byte)(
            (
                (hi >= '0' && hi <= '9') ? hi - '0' :
                (hi >= 'a' && hi <= 'f') ? hi - 'a' + 10 :
                (hi >= 'A' && hi <= 'F') ? hi - 'A' + 10 :
                0
            )
            << 4 | 
            (
                (lo >= '0' && lo <= '9') ? lo - '0' :
                (lo >= 'a' && lo <= 'f') ? lo - 'a' + 10 :
                (lo >= 'A' && lo <= 'F') ? lo - 'A' + 10 :
                0
            )
        );
    }

    return rc;          
}

这是一篇很棒的帖子。我喜欢瓦利德的解决方案。我还没有通过帕特里奇的测试,但似乎很快。我还需要反向过程,将十六进制字符串转换为字节数组,因此我将其作为Waleed解决方案的反向来编写。不确定它是否比托马拉克的原始解决方案更快。同样,我也没有通过帕特里奇的测试运行相反的过程。

private byte[] HexStringToByteArray(string hexString)
{
    int hexStringLength = hexString.Length;
    byte[] b = new byte[hexStringLength / 2];
    for (int i = 0; i < hexStringLength; i += 2)
    {
        int topChar = (hexString[i] > 0x40 ? hexString[i] - 0x37 : hexString[i] - 0x30) << 4;
        int bottomChar = hexString[i + 1] > 0x40 ? hexString[i + 1] - 0x37 : hexString[i + 1] - 0x30;
        b[i / 2] = Convert.ToByte(topChar + bottomChar);
    }
    return b;
}

具有扩展支持的基本解决方案

public static class Utils
{
    public static byte[] ToBin(this string hex)
    {
        int NumberChars = hex.Length;
        byte[] bytes = new byte[NumberChars / 2];
        for (int i = 0; i < NumberChars; i += 2)
            bytes[i / 2] = Convert.ToByte(hex.Substring(i, 2), 16);
        return bytes;
    }
    public static string ToHex(this byte[] ba)
    {
        return  BitConverter.ToString(ba).Replace("-", "");
    }
}

并像下面那样使用这个类

    byte[] arr1 = new byte[] { 1, 2, 3 };
    string hex1 = arr1.ToHex();
    byte[] arr2 = hex1.ToBin();