将二进制长字符串转换为十六进制c#

我刚刚搞砸了。也许你可以把。。。

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();
}

这可能会对您有所帮助：

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

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

对于长于此的字符串，您可以简单地将其分解为多个字节：

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")));

我想出了这个方法。我是编程和C#的新手，但我希望你会喜欢它：

static string BinToHex(string bin)
{
    StringBuilder binary = new StringBuilder(bin);
    bool isNegative = false;
    if (binary[0] == '-')
    {
        isNegative = true;
        binary.Remove(0, 1);
    }
    for (int i = 0, length = binary.Length; i < (4 - length % 4) % 4; i++) //padding leading zeros
    {
        binary.Insert(0, '0');
    }
    StringBuilder hexadecimal = new StringBuilder();
    StringBuilder word = new StringBuilder("0000");
    for (int i = 0; i < binary.Length; i += 4)
    {
        for (int j = i; j < i + 4; j++)
        {
            word[j % 4] = binary[j];
        }
        switch (word.ToString())
        {
            case "0000": hexadecimal.Append('0'); break;
            case "0001": hexadecimal.Append('1'); break;
            case "0010": hexadecimal.Append('2'); break;
            case "0011": hexadecimal.Append('3'); break;
            case "0100": hexadecimal.Append('4'); break;
            case "0101": hexadecimal.Append('5'); break;
            case "0110": hexadecimal.Append('6'); break;
            case "0111": hexadecimal.Append('7'); break;
            case "1000": hexadecimal.Append('8'); break;
            case "1001": hexadecimal.Append('9'); break;
            case "1010": hexadecimal.Append('A'); break;
            case "1011": hexadecimal.Append('B'); break;
            case "1100": hexadecimal.Append('C'); break;
            case "1101": hexadecimal.Append('D'); break;
            case "1110": hexadecimal.Append('E'); break;
            case "1111": hexadecimal.Append('F'); break;
            default:
                return "Invalid number";
        }
    }
    if (isNegative)
    {
        hexadecimal.Insert(0, '-');
    }
    return hexadecimal.ToString();
}

考虑到四个位可以用一个十六进制值表示，您可以简单地按四个一组进行转换，值不会以这种方式改变。

string bin = "11110110";
int rest = bin.Length % 4;
if(rest != 0)
    bin = new string('0', 4-rest) + bin; //pad the length out to by divideable by 4
string output = "";
for(int i = 0; i <= bin.Length - 4; i +=4)
{
    output += string.Format("{0:X}", Convert.ToByte(bin.Substring(i, 4), 2));
}

如果您想迭代字符串中每个字节的十六进制表示，可以使用以下扩展。我把米奇的回答和这个结合起来了。

static class StringExtensions
{
    public static IEnumerable<string> ToHex(this String s) {
        if (s == null)
            throw new ArgumentNullException("s");
        int mod4Len = s.Length % 8;
        if (mod4Len != 0)
        {
            // pad to length multiple of 8
            s = s.PadLeft(((s.Length / 8) + 1) * 8, '0');
        }
        int numBitsInByte = 8;
        for (var i = 0; i < s.Length; i += numBitsInByte)
        {
            string eightBits = s.Substring(i, numBitsInByte);
            yield return string.Format("{0:X2}", Convert.ToByte(eightBits, 2));
        }
    }
}

示例：

string test = "0110011010010111001001110101011100110100001101101000011001010110001101101011";
foreach (var hexVal in test.ToHex())
{
    Console.WriteLine(hexVal);  
}

打印

06
69
72
75
73
43
68
65
63
6B

如果您使用.NET 4.0或更高版本，并且愿意使用System.Numerics.dll(用于BigInteger类(，则以下解决方案可以正常工作：

public static string ConvertBigBinaryToHex(string bigBinary)
{
    BigInteger bigInt = BigInteger.Zero;
    int exponent = 0;
    for (int i = bigBinary.Length - 1; i >= 0; i--, exponent++)
    {
        if (bigBinary[i] == '1')
            bigInt += BigInteger.Pow(2, exponent);
    }
    return bigInt.ToString("X");
}

考虑到四个位可以用一个十六进制值表示，您可以简单地按四个一组进行转换，值不会以这种方式改变。

string bin = "11110110";
int rest = bin.Length % 4;
bin = bin.PadLeft(rest, '0'); //pad the length out to by divideable by 4
string output = "";
for(int i = 0; i <= bin.Length - 4; i +=4)
{
    output += string.Format("{0:X}", Convert.ToByte(bin.Substring(i, 4), 2));
}

static string BinToHex(string bin)
{
    if (bin == null)
        throw new ArgumentNullException("bin");
    if (bin.Length % 8 != 0)
        throw new ArgumentException("The length must be a multiple of 8", "bin");
    var hex = Enumerable.Range(0, bin.Length / 8)
                     .Select(i => bin.Substring(8 * i, 8))
                     .Select(s => Convert.ToByte(s, 2))
                     .Select(b => b.ToString("x2"));
    return String.Join(null, hex);
}

使用LINQ

   string BinaryToHex(string binaryString)
            {
                var offset = 0;
                StringBuilder sb = new();
                
                while (offset < binaryString.Length)
                {
                    var nibble = binaryString
                        .Skip(offset)
                        .Take(4);
                    sb.Append($"{Convert.ToUInt32(nibble.toString()), 2):X}");
                    offset += 4;
                }
                return sb.ToString();
            }

您可以一次输入四位数字。将这个数字转换为ex(正如您所做的那样(，然后将字符串连接在一起。因此，您可以获得一个以十六进制表示数字的字符串，与大小无关。根据输入字符串的MSB起始位置，可能是您获得的输出字符串，我所描述的方式必须反转。