将长字符串转换为十六进制c#

Question

我正在寻找一种将长字符串转换为十六进制字符串的方法.

二进制字符串看起来像这样 "0110011010010111001001110101011100110100001101101000011001010110001101101011"

我试过用了

hex = String.Format("{0:X2}", Convert.ToUInt64(hex, 2));

但这只适用于二进制字符串适合Uint64的情况,如果字符串足够长则不会.

有另一种方法将二进制字符串转换为十六进制？

谢谢

Answer 1

我刚敲了这个.也许你可以用作起点......

public static string BinaryStringToHexString(string binary)
{
    if (string.IsNullOrEmpty(binary))
        return binary;

    StringBuilder result = new StringBuilder(binary.Length / 8 + 1);

    // TODO: check all 1's or 0's... throw otherwise

    int mod4Len = binary.Length % 8;
    if (mod4Len != 0)
    {
        // pad to length multiple of 8
        binary = binary.PadLeft(((binary.Length / 8) + 1) * 8, '0');
    }

    for (int i = 0; i < binary.Length; i += 8)
    {
        string eightBits = binary.Substring(i, 8);
        result.AppendFormat("{0:X2}", Convert.ToByte(eightBits, 2));
    }

    return result.ToString();
}

Answer 2

这可能对您有所帮助:

string HexConverted(string strBinary)
    {
        string strHex = Convert.ToInt32(strBinary,2).ToString("X");
        return strHex;
    }

Answer 3

Convert.ToInt32("1011", 2).ToString("X");

对于比此更长的字符串,您可以简单地将其分成多个字节:

var binary = "0110011010010111001001110101011100110100001101101000011001010110001101101011";
var hex = string.Join(" ", 
            Enumerable.Range(0, binary.Length / 8)
            .Select(i => Convert.ToByte(binary.Substring(i * 8, 8), 2).ToString("X2")));