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));

The other way is to use BinaryPrimitives like so

byte[] intBytes = BitConverter.GetBytes(123); int actual = BinaryPrimitives.ReadInt32LittleEndian(intBytes);

Why all this code in the samples above...

A struct with explicit layout acts both ways and has no performance hit.

Update: Since there's a question on how to deal with endianness I added an interface that illustrates how to abstract that. Another implementing struct can deal with the opposite case

public interface IIntToByte
{
    Int32 Int { get; set;}

    byte B0 { get; }
    byte B1 { get; }
    byte B2 { get; }
    byte B3 { get; }
}

[StructLayout(LayoutKind.Explicit)]
public struct IntToByteLE : UserQuery.IIntToByte
{
    [FieldOffset(0)]
    public Int32 IntVal;

    [FieldOffset(0)]
    public byte b0;
    [FieldOffset(1)]
    public byte b1;
    [FieldOffset(2)]
    public byte b2;
    [FieldOffset(3)]
    public byte b3;

    public Int32 Int {
        get{ return IntVal; }
        set{ IntVal = value;}
    }

    public byte B0 => b0;
    public byte B1 => b1;
    public byte B2 => b2;
    public byte B3 => b3; 
}

When I look at this description, I have a feeling, that this xdr integer is just a big-endian "standard" integer, but it's expressed in the most obfuscated way. Two's complement notation is better know as U2, and it's what we are using on today's processors. The byte order indicates that it's a big-endian notation.
So, answering your question, you should inverse elements in your array (0 <--> 3, 1 <-->2), as they are encoded in little-endian. Just to make sure, you should first check BitConverter.IsLittleEndian to see on what machine you are running.

using static System.Console;

namespace IntToBits
{
    class Program
    {
        static void Main()
        {
            while (true)
            {
                string s = Console.ReadLine();
                Clear();
                uint i;
                bool b = UInt32.TryParse(s, out i);
                if (b) IntPrinter(i);
            }
        }

        static void IntPrinter(uint i)
        {
            int[] iarr = new int [32];
            Write("[");
            for (int j = 0; j < 32; j++)
            {
                uint tmp = i & (uint)Math.Pow(2, j);

                iarr[j] = (int)(tmp >> j);
            }
            for (int j = 32; j > 0; j--)
            {
                if(j%8==0 && j != 32)Write("|");
                if(j%4==0 && j%8 !=0) Write("'");
                Write(iarr[j-1]);
            }
            WriteLine("]");
        }
    }
}```

I wouldn’t recommend using byte arrays to get bytes from integers stored in the stack. It produces an array that will be stored in the heap until the next GC clean up. I prefer using spans for these simple tasks.

So my solution is here to use MemoryMarshal.AsBytes<T>(ReadOnlySpan<T>). You don't need to worry about the endianness, about casting. You just read the method name and understand that this will return bytes from any type of span. You just wrap your integer into a span via MemoryMarshal.CreateReadOnlySpan<T>(T, Int32). Those bytes will be stored in the stack and terminated after leaving the scope.

Full solution will be:

var integer = int.MaxValue;

var readOnlySpan = MemoryMarshal.CreateReadOnlySpan(ref integer, length: 1);
var bytes = MemoryMarshal.AsBytes(readOnlySpan);

Note that AsBytes provides raw representation of the value. You can accidentally change the span and it will be reflected on the integer itself. So don't forget to always use read-only spans when you don't want to change anything inside.

byte[] Take_Byte_Arr_From_Int(Int64 Source_Num)
{
   Int64 Int64_Num = Source_Num;
   byte Byte_Num;
   byte[] Byte_Arr = new byte[8];
   for (int i = 0; i < 8; i++)
   {
      if (Source_Num > 255)
      {
         Int64_Num = Source_Num / 256;
         Byte_Num = (byte)(Source_Num - Int64_Num * 256);
      }
      else
      {
         Byte_Num = (byte)Int64_Num;
         Int64_Num = 0;
      }
      Byte_Arr[i] = Byte_Num;
      Source_Num = Int64_Num;
   }
   return (Byte_Arr);
}