C#, BitConverter.ToUInt32, incorrect value

Question

I try to convert an Ip Address to a long value :

byte[] Ip = new byte[4] { 192, 168, 1, 0 };

UInt32 Ret1 = (((UInt32)Ip[0]) << 24) |
              (((UInt32)Ip[1]) << 16) |
              (((UInt32)Ip[2]) << 8)  |
              (((UInt32)Ip[3]));

UInt32 Ret2 = BitConverter.ToUInt32(Ip, 0);

Ret1 returns 3232235776 (the correct value)

Ret2 returns 108736 (?)

Why this difference ?

Answer 1

Indeed, endianness is your issue here. While not difficult to work around, it can be annoying at times on Intel-based systems due to them being in Little Endian whereas Network Order is Big Endian. In short, your bytes are in the reverse order.

Here is a little convenience method you may use to solve this issue (and even on various platforms):

static uint MakeIPAddressInt32(byte[] array)
{
  // Make a defensive copy.
  var ipBytes = new byte[array.Length];
  array.CopyTo(ipBytes, 0);

  // Reverse if we are on a little endian architecture.
  if(BitConverter.IsLittleEndian)
    Array.Reverse(ipBytes);

  // Convert these bytes to an unsigned 32-bit integer (IPv4 address).
  return BitConverter.ToUInt32(ipBytes, 0);
} 

Answer 2

Sounds like you have a problem with endianness.

http://en.wikipedia.org/wiki/Endianness

Answer 3

The platform (bitness) independent code can looks lake this:

UInt32 Ret2 = BitConverter.ToUInt32(
                   BitConverter.IsLittleEndian
                   ? Ip.Reverse().ToArray()
                   : Ip, 0);

Answer 4

It's an endian issue. IPAddress.HostToNetworkOrder has various overloads for converting numeric types to their network equivilant (ie little endian to big endian).

Answer 5

IP addresses are typically written in Network Byte Order which happens to be the other endianness of standard x86.

Google for aton(), ntoa() and big endian.