Converting long string of binary to hex c#
I'm looking for a way to convert a long string of binary to a hex string.
开发者_JAVA技巧the binary string looks something like this "0110011010010111001001110101011100110100001101101000011001010110001101101011"
I've tried using
hex = String.Format("{0:X2}", Convert.ToUInt64(hex, 2));
but that only works if the binary string fits into a Uint64 which if the string is long enough it won't.
is there another way to convert a string of binary into hex?
Thanks
I just knocked this up. Maybe you can use as a starting point...
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();
}
This might help you:
string HexConverted(string strBinary)
{
string strHex = Convert.ToInt32(strBinary,2).ToString("X");
return strHex;
}
Convert.ToInt32("1011", 2).ToString("X");
For string longer than this, you can simply break it into multiple bytes:
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")));
I came up with this method. I am new to programming and C# but I hope you will appreciate it:
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();
}
Considering four bits can be expressed by one hex value, you can simply go by groups of four and convert them seperately, the value won't change that way.
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));
}
If you want to iterate over the hexadecimal representation of each byte in the string, you could use the following extension. I've combined Mitch's answer with this.
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));
}
}
}
Example:
string test = "0110011010010111001001110101011100110100001101101000011001010110001101101011";
foreach (var hexVal in test.ToHex())
{
Console.WriteLine(hexVal);
}
Prints
06
69
72
75
73
43
68
65
63
6B
If you're using .NET 4.0 or later and if you're willing to use System.Numerics.dll (for BigInteger class), the following solution works fine:
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");
}
Considering four bits can be expressed by one hex value, you can simply go by groups of four and convert them seperately, the value won't change that way.
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);
}
Using 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();
}
You can take the input number four digit at a time. Convert this digit to ex ( as you did is ok ) then concat the string all together. So you obtain a string representing the number in hex, independetly from the size. Depending on where start MSB on your input string, may be the output string you obtain the way i described must be reversed.
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