How to convert an IPv4 address into a integer in C#?

I'm looking for a function that will convert a standard IPv4 address into an Integer. Bonus points available for a function that will do the opposite.

Solution should be in C#.

32-bit unsigned integers are IPv4 addresses. Meanwhile, the IPAddress.Address property, while deprecated, is an Int64 that returns the unsigned 32-bit value of the IPv4 address (the catch is, it's in network byte order, so you need to swap it around).

For example, my local google.com is at 64.233.187.99. That's equivalent to:

64*2^24 + 233*2^16 + 187*2^8 + 99
= 1089059683

And indeed, http://1089059683/ works as expected (at least in Windows, tested with IE, Firefox and Chrome; doesn't work on iPhone though).

Here's a test program to show both conversions, including the network/host byte swapping:

using System;
using System.Net;

class App
{
    static long ToInt(string addr)
    {
        // careful of sign extension: convert to uint first;
        // unsigned NetworkToHostOrder ought to be provided.
        return (long) (uint) IPAddress.NetworkToHostOrder(
             (int) IPAddress.Parse(addr).Address);
    }

    static string ToAddr(long address)
    {
        return IPAddress.Parse(address.ToString()).ToString();
        // This also works:
        // return new IPAddress((uint) IPAddress.HostToNetworkOrder(
        //    (int) address)).ToString();
    }

    static void Main()
    {
        Console.WriteLine(ToInt("64.233.187.99"));
        Console.WriteLine(ToAddr(1089059683));
    }
}

Here's a pair of methods to convert from IPv4 to a correct integer and back:

public static uint ConvertFromIpAddressToInteger(string ipAddress)
{
    var address = IPAddress.Parse(ipAddress);
    byte[] bytes = address.GetAddressBytes();

    // flip big-endian(network order) to little-endian
    if (BitConverter.IsLittleEndian)
    {
        Array.Reverse(bytes);
    }

    return BitConverter.ToUInt32(bytes, 0);
}

public static string ConvertFromIntegerToIpAddress(uint ipAddress)
{
    byte[] bytes = BitConverter.GetBytes(ipAddress);

    // flip little-endian to big-endian(network order)
    if (BitConverter.IsLittleEndian)
    {
        Array.Reverse(bytes);
    }

    return new IPAddress(bytes).ToString();
}

Example

ConvertFromIpAddressToInteger("255.255.255.254"); // 4294967294
ConvertFromIntegerToIpAddress(4294967294); // 255.255.255.254

Explanation

IP addresses are in network order (big-endian), while ints are little-endian on Windows, so to get a correct value, you must reverse the bytes before converting on a little-endian system.

Also, even for IPv4, an int can't hold addresses bigger than 127.255.255.255, e.g. the broadcast address (255.255.255.255), so use a uint.

@Barry Kelly and @Andrew Hare, actually, I don't think multiplying is the most clear way to do this (alltough correct).

An Int32 "formatted" IP address can be seen as the following structure

[StructLayout(LayoutKind.Sequential, Pack = 1)] 
struct IPv4Address
{
   public Byte A;
   public Byte B;
   public Byte C;
   public Byte D;
} 
// to actually cast it from or to an int32 I think you 
// need to reverse the fields due to little endian

So to convert the ip address 64.233.187.99 you could do:

(64  = 0x40) << 24 == 0x40000000
(233 = 0xE9) << 16 == 0x00E90000
(187 = 0xBB) << 8  == 0x0000BB00
(99  = 0x63)       == 0x00000063
                      ---------- =|
                      0x40E9BB63

so you could add them up using + or you could binairy or them together. Resulting in 0x40E9BB63 which is 1089059683. (In my opinion looking in hex it's much easier to see the bytes)

So you could write the function as:

int ipToInt(int first, int second, 
    int third, int fourth)
{
    return (first << 24) | (second << 16) | (third << 8) | (fourth);
}