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

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();

其他回答

Dotnet 5更新

要从byte[](字节数组)转换为十六进制字符串,请使用:

System.Convert.ToHexString

var myBytes = new byte[100];
var myString = System.Convert.ToHexString(myBytes);

要将十六进制字符串转换为字节[],请使用:

System.Convert.FromHexString

var myString  = "E10B116E8530A340BCC7B3EAC208487B";
var myBytes = System.Convert.FromHexString(myString);

就速度而言,这似乎比这里的任何东西都要好:

  public static string ToHexString(byte[] data) {
    byte b;
    int i, j, k;
    int l = data.Length;
    char[] r = new char[l * 2];
    for (i = 0, j = 0; i < l; ++i) {
      b = data[i];
      k = b >> 4;
      r[j++] = (char)(k > 9 ? k + 0x37 : k + 0x30);
      k = b & 15;
      r[j++] = (char)(k > 9 ? k + 0x37 : k + 0x30);
    }
    return new string(r);
  }

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("-", "")

我猜它的速度值16个额外的字节。

    static char[] hexes = new char[]{'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
    public static string ToHexadecimal (this byte[] Bytes)
    {
        char[] Result = new char[Bytes.Length << 1];
        int Offset = 0;
        for (int i = 0; i != Bytes.Length; i++) {
            Result[Offset++] = hexes[Bytes[i] >> 4];
            Result[Offset++] = hexes[Bytes[i] & 0x0F];
        }
        return new string(Result);
    }

这里不想赘述很多答案,但我发现了一个十六进制字符串解析器的相当优化(比公认的好4.5倍)、简单的实现。首先,我的测试输出(第一批是我的实现):

Give me that string:
04c63f7842740c77e545bb0b2ade90b384f119f6ab57b680b7aa575a2f40939f

Time to parse 100,000 times: 50.4192 ms
Result as base64: BMY/eEJ0DHflRbsLKt6Qs4TxGfarV7aAt6pXWi9Ak58=
BitConverter'd: 04-C6-3F-78-42-74-0C-77-E5-45-BB-0B-2A-DE-90-B3-84-F1-19-F6-AB-5
7-B6-80-B7-AA-57-5A-2F-40-93-9F

Accepted answer: (StringToByteArray)
Time to parse 100000 times: 233.1264ms
Result as base64: BMY/eEJ0DHflRbsLKt6Qs4TxGfarV7aAt6pXWi9Ak58=
BitConverter'd: 04-C6-3F-78-42-74-0C-77-E5-45-BB-0B-2A-DE-90-B3-84-F1-19-F6-AB-5
7-B6-80-B7-AA-57-5A-2F-40-93-9F

With Mono's implementation:
Time to parse 100000 times: 777.2544ms
Result as base64: BMY/eEJ0DHflRbsLKt6Qs4TxGfarV7aAt6pXWi9Ak58=
BitConverter'd: 04-C6-3F-78-42-74-0C-77-E5-45-BB-0B-2A-DE-90-B3-84-F1-19-F6-AB-5
7-B6-80-B7-AA-57-5A-2F-40-93-9F

With SoapHexBinary:
Time to parse 100000 times: 845.1456ms
Result as base64: BMY/eEJ0DHflRbsLKt6Qs4TxGfarV7aAt6pXWi9Ak58=
BitConverter'd: 04-C6-3F-78-42-74-0C-77-E5-45-BB-0B-2A-DE-90-B3-84-F1-19-F6-AB-5
7-B6-80-B7-AA-57-5A-2F-40-93-9F

base64和“BitConverter'd”行用于测试正确性。请注意,它们是相等的。

实施:

public static byte[] ToByteArrayFromHex(string hexString)
{
  if (hexString.Length % 2 != 0) throw new ArgumentException("String must have an even length");
  var array = new byte[hexString.Length / 2];
  for (int i = 0; i < hexString.Length; i += 2)
  {
    array[i/2] = ByteFromTwoChars(hexString[i], hexString[i + 1]);
  }
  return array;
}

private static byte ByteFromTwoChars(char p, char p_2)
{
  byte ret;
  if (p <= '9' && p >= '0')
  {
    ret = (byte) ((p - '0') << 4);
  }
  else if (p <= 'f' && p >= 'a')
  {
    ret = (byte) ((p - 'a' + 10) << 4);
  }
  else if (p <= 'F' && p >= 'A')
  {
    ret = (byte) ((p - 'A' + 10) << 4);
  } else throw new ArgumentException("Char is not a hex digit: " + p,"p");

  if (p_2 <= '9' && p_2 >= '0')
  {
    ret |= (byte) ((p_2 - '0'));
  }
  else if (p_2 <= 'f' && p_2 >= 'a')
  {
    ret |= (byte) ((p_2 - 'a' + 10));
  }
  else if (p_2 <= 'F' && p_2 >= 'A')
  {
    ret |= (byte) ((p_2 - 'A' + 10));
  } else throw new ArgumentException("Char is not a hex digit: " + p_2, "p_2");

  return ret;
}

我尝试了一些不安全的东西,并将(显然是冗余的)字符移动到另一个方法来蚕食if序列,但这是最快的。

(我承认这回答了一半的问题。我觉得字符串->字节[]转换不足,而字节[]->字符串角度似乎被很好地覆盖了。因此,这个答案。)