C# int to byte[]

The RFC is just trying to say that a signed integer is a normal 4-byte integer with bytes ordered in a big-endian way.

Now, you are most probably working on a little-endian machine and BitConverter.GetBytes() will give you the byte[] reversed. So you could try:

int intValue;
byte[] intBytes = BitConverter.GetBytes(intValue);
Array.Reverse(intBytes);
byte[] result = intBytes;

For the code to be most portable, however, you can do it like this:

int intValue;
byte[] intBytes = BitConverter.GetBytes(intValue);
if (BitConverter.IsLittleEndian)
    Array.Reverse(intBytes);
byte[] result = intBytes;

Here's another way to do it: as we all know 1x byte = 8x bits and also, a "regular" integer (int32) contains 32 bits (4 bytes). We can use the >> operator to shift bits right (>> operator does not change value.)

int intValue = 566;

byte[] bytes = new byte[4];

bytes[0] = (byte)(intValue >> 24);
bytes[1] = (byte)(intValue >> 16);
bytes[2] = (byte)(intValue >> 8);
bytes[3] = (byte)intValue;

Console.WriteLine("{0} breaks down to : {1} {2} {3} {4}",
    intValue, bytes[0], bytes[1], bytes[2], bytes[3]);

BitConverter.GetBytes(int) almost does what you want, except the endianness is wrong.

You can use the IPAddress.HostToNetwork method to swap the bytes within the the integer value before using BitConverter.GetBytes or use Jon Skeet's EndianBitConverter class. Both methods do the right thing(tm) regarding portability.

int value;
byte[] bytes = BitConverter.GetBytes(IPAddress.HostToNetworkOrder(value));