将十六进制字符串字节转换为字节数组

Question

I am building a client application which needs to send packet to server in a particular format such that if the byte is 0xA3, the server expects it as {0x3A, 0x33}

I had used the below approach earlier. It works well if the byte is for instance 0x89. But if the byte is 0xA3 it doesnt work

string hex = hexStr .Length == 1 ? "0" + hexStr:hexStr ;
byte packet1 = (byte)(int.Parse(hex[0].ToString(), System.Globalization.NumberStyles.HexNumber) + 0x30);
byte packet2 = (byte)(int.Parse(hex[1].ToString(), System.Globalization.NumberStyles.HexNumber) + 0x30);

Examples of expected output

1. input => 0x89 , Output => {0x38, 0x39}
2. input => 0xA3 , Output => {0x3A, 0x33}

However if i use the above code i get the following output

1. input => 0x89 , Output => {0x38, 0x39}
2. input => 0xA3 , Output => {0x41, 0x33}

Answer 1

The problem isn't in the code you've shown.

You need to do all your math in hexadecimal, and convert to strings (if needed) into hexadecimal as well:

string hex = "A3";
byte packet1 = (byte)(int.Parse(hex[0].ToString(), NumberStyles.HexNumber) + 0x30);
byte packet2 = (byte)(int.Parse(hex[1].ToString(), NumberStyles.HexNumber) + 0x30);

Console.WriteLine("{0:X2}, {1:X2}", packet1, packet2); // 3A, 33

works exactly as you expect.

The results you got seem to indicate that you took 30 (decimal), added 0xA (11), and printed as decimal, rather than hexa-decimal. This does not happen in the code you've posted, so just fix your actual code and you'll be fine.