How can I convert a hex string to a byte array? [duplicate]

Solution 1:

Here's a nice fun LINQ example.

public static byte[] StringToByteArray(string hex) {
    return Enumerable.Range(0, hex.Length)
                     .Where(x => x % 2 == 0)
                     .Select(x => Convert.ToByte(hex.Substring(x, 2), 16))
                     .ToArray();
}

Solution 2:

I did some research and found out that byte.Parse is even slower than Convert.ToByte. The fastest conversion I could come up with uses approximately 15 ticks per byte.

    public static byte[] StringToByteArrayFastest(string hex) {
        if (hex.Length % 2 == 1)
            throw new Exception("The binary key cannot have an odd number of digits");

        byte[] arr = new byte[hex.Length >> 1];

        for (int i = 0; i < hex.Length >> 1; ++i)
        {
            arr[i] = (byte)((GetHexVal(hex[i << 1]) << 4) + (GetHexVal(hex[(i << 1) + 1])));
        }

        return arr;
    }

    public static int GetHexVal(char hex) {
        int val = (int)hex;
        //For uppercase A-F letters:
        //return val - (val < 58 ? 48 : 55);
        //For lowercase a-f letters:
        //return val - (val < 58 ? 48 : 87);
        //Or the two combined, but a bit slower:
        return val - (val < 58 ? 48 : (val < 97 ? 55 : 87));
    }

// also works on .NET Micro Framework where (in SDK4.3) byte.Parse(string) only permits integer formats.

Solution 3:

The following code changes the hexadecimal string to a byte array by parsing the string byte-by-byte.

public static byte[] ConvertHexStringToByteArray(string hexString)
{
    if (hexString.Length % 2 != 0)
    {
        throw new ArgumentException(String.Format(CultureInfo.InvariantCulture, "The binary key cannot have an odd number of digits: {0}", hexString));
    }

    byte[] data = new byte[hexString.Length / 2];
    for (int index = 0; index < data.Length; index++)
    {
        string byteValue = hexString.Substring(index * 2, 2);
        data[index] = byte.Parse(byteValue, NumberStyles.HexNumber, CultureInfo.InvariantCulture);
    }

    return data; 
}