13

I'm trying to parse a string containing an IP address and a port using IPAddress.Parse. This works well with IPv6 addresses but not with IPv4 addresses. Can somone explain why this happens?

The code I'm using is:

IPAddress.Parse("[::1]:5"); //Valid
IPAddress.Parse("127.0.0.1:5"); //null
abatishchev
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Francisco Silva
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    Here I don't get a null for the second example. I get an exception. An IPv4 addresses does not include port numbers. Note you're trying to parse a string containing "an IP address *and* a port". So the exception makes sense. Why the IPv6 version works, I don't know. – R. Martinho Fernandes Feb 11 '11 at 12:20
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    `IPAddress.Parse("[::1]:5");` is valid, *but* the `:5` is silently dropped! If you inspect the resulting object, you can see that the result is just `::1`. This might actually be a bug in the `IPAddress.Parse` method ... – hangy Feb 11 '11 at 12:29
  • @Martinho Fernandes - it's not a bug in the IP6 parser, [::1]:5 is a valid IP6 address because the separator is ':', not '::' (http://tools.ietf.org/html/draft-ietf-6man-text-addr-representation-04#section-4) - **and** combining Ports is built into the IP6 spec as well, however the IPAddress entity itself is not expected to *contain* the port. It's possible that [a,b,c,d]:port for ipv4 is also allowed since the IP6 spec seems to suggest it was part of the IP4 spec – Andras Zoltan Feb 11 '11 at 12:46
  • @Andras: `::1:5` is indeed a valid IPv6 address, but not the same thing as `[::1]:5`, which is port 5 on address `::1`. Since an "IP address" is not an "IP address and a port", I think an exception would be the correct behavior, at least to match with IPv4 (btw [a.b.c.d]:port throws as well). The port is a part of the IPv6 protocol, but not a part of an IPv6 address. – R. Martinho Fernandes Feb 11 '11 at 12:51
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    @Martinho Fernandes- Actually I meant to direct the comment at @hangy, sorry! :$-to be a pedant, it is a valid *text representation* for IPv6. If you Read the IPv6 RFC, it specifically allows for IP6 addresses encoded with a port as part of the *text representation*. It doesn't say that the port must be retained (obviously it can't) - but it defines how it should be handled when parsing. The irony being that if the IPv4 spec also allows for this, then MS should update the IPv4 code to support it also; or not have written the IPv6 parser to support it, rather than doing a half-baked job :) – Andras Zoltan Feb 11 '11 at 13:00
  • @Andras: Are you referring to the section titled ["6. Notes on Combining IPv6 Addresses with Port Numbers"](http://tools.ietf.org/html/draft-ietf-6man-text-addr-representation-04#section-6)? That refers to "IPv6 addresses and port numbers are represented in text *combined* together" (emphasis mine). Or am I missing something? – R. Martinho Fernandes Feb 11 '11 at 13:04
  • @Martinho Fernandes - so I guess I'm also agreeing with you there at we should at least have some consistency from the two scenarios! – Andras Zoltan Feb 11 '11 at 13:06
  • @Martinho Fernandes - yes, and - anticipating your next comment - I guess it could be argued successfully that it's not actually part of the RFC, but a side-note; fair point. Seems like the .Net software engineers have read that and implemented it 'to be helpful'; hmmm... they probably shouldn't have done. – Andras Zoltan Feb 11 '11 at 13:09
  • @Martinho Fernandes - or, to put it another way. I was wrong :)! – Andras Zoltan Feb 11 '11 at 13:27
  • So one could argue that the IPv6 behaviour actually is a bug and should fail in the same way as the IPv4 does. It may be interesting to note that Mono behaves different to .NET Framework: http://ideone.com/CRrW9 – hangy Feb 13 '11 at 14:40

7 Answers7

23
Uri url;
IPAddress ip;
if (Uri.TryCreate(String.Format("http://{0}", "127.0.0.1:5"), UriKind.Absolute, out url) &&
   IPAddress.TryParse(url.Host, out ip))
{
    IPEndPoint endPoint = new IPEndPoint(ip, url.Port);
}
abatishchev
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11

This happens because the port is not part of the IP address. It belongs to TCP/UDP, and you'll have to strip it out first. The Uri class might be helpful for this.

Alex J
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7

IPAddress is not IP+Port. You want IPEndPoint.

Example from http://www.java2s.com/Code/CSharp/Network/ParseHostString.htm

public static void ParseHostString(string hostString, ref string hostName, ref int port)
{
   hostName = hostString;
   if (hostString.Contains(":"))
   {
      string[] hostParts = hostString.Split(':');

      if (hostParts.Length == 2)
      {
         hostName = hostParts[0];
         int.TryParse(hostParts[1], out port);
      }
   }
}

Edit: Ok, I'll admit that wasn't the most elegant solution. Try this one I wrote (just for you) instead:

// You need to include some usings:
using System.Text.RegularExpressions;
using System.Net;

// Then this code (static is not required):
private static Regex hostPortMatch = new Regex(@"^(?<ip>(?:\[[\da-fA-F:]+\])|(?:\d{1,3}\.){3}\d{1,3})(?::(?<port>\d+))?$", System.Text.RegularExpressions.RegexOptions.Compiled);
public static IPEndPoint ParseHostPort(string hostPort)
{
   Match match = hostPortMatch.Match(hostPort);
   if (!match.Success)
      return null;

   return new IPEndPoint(IPAddress.Parse(match.Groups["ip"].Value), int.Parse(match.Groups["port"].Value));
}

Note that this one ONLY accepts IP address, not hostname. If you want to support hostname you'll either have to resolve it to IP or not use IPAddress/IPEndPoint.

Koen Zomers
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Tedd Hansen
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  • Thanks for your effort, but I had already come up with a solution for my problem using the Uri class. I was just intrigued as to why the behavior of IPAddress.Parse in these two situations. – Francisco Silva Feb 11 '11 at 18:50
  • Your 2nd code example's RegEx doesn't even match the most basic of endpoints, `127.0.0.1:80`. – keeehlan Oct 23 '13 at 22:16
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    the first solution is the best. it doesn't matter if it doesn't _look_ clever. – Boppity Bop Dec 08 '18 at 12:49
5

IPAddress.Parse is meant to take A string that contains an IP address in dotted-quad notation for IPv4 and in colon-hexadecimal notation for IPv6. So your first example works for IPv6 and your second example fails because it doesnt support a port for IPv4. Link http://msdn.microsoft.com/en-us/library/system.net.ipaddress.parse.aspx

David Yenglin
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3

As Tedd Hansen pointed out, what you are trying to parse is not an IP address but an IP endpoint (IP address + port). And since .NET Core 3.0, you can use IPEndPoint.TryParse to parse a string as an IPEndPoint:

if (IPEndPoint.TryParse("127.0.0.1:5", out IPEndPoint endpoint))
{
    // parsed successfully, you can use the "endpoint" variable
    Console.WriteLine(endpoint.Address.ToString()); // writes "127.0.0.1"
    Console.WriteLine(endpoint.Port.ToString()); // writes "5"
}
else
{
    // failed to parse
}
0xced
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2

If you work on older versions of .net you can take IPEndPoint.Parse implementation from open source: https://github.com/dotnet/runtime/blob/main/src/libraries/System.Net.Primitives/src/System/Net/IPEndPoint.cs

Yuriy Gavrishov
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0

To add my two cents... Since Microsoft itself implemented TryParse in NET Core 3.0 I've opted to stop using my custom IP+Port parser and kindly borrowed their code with some adaptations:

public static class IPEndPointParserExtension
{
    public static bool TryParseAsIPEndPoint(this string s, out IPEndPoint result) {
#if NETCOREAPP3_0_OR_GREATER
        return IPEndPoint.TryParse(s, out result);
#else
        int addressLength = s.Length;  // If there's no port then send the entire string to the address parser
        int lastColonPos = s.LastIndexOf(':');
        // Look to see if this is an IPv6 address with a port.
        if (lastColonPos > 0) {
            if (s[lastColonPos - 1] == ']') 
                addressLength = lastColonPos;
            // Look to see if this is IPv4 with a port (IPv6 will have another colon)
            else if (s.Substring(0, lastColonPos).LastIndexOf(':') == -1) 
                addressLength = lastColonPos;
        }
        if (IPAddress.TryParse(s.Substring(0, addressLength), out IPAddress address)) {
            long port = 0;
            if (addressLength == s.Length ||
                (long.TryParse(s.Substring(addressLength + 1), out port) 
                    && port <= IPEndPoint.MaxPort)) {
                result = new IPEndPoint(address, (int)port);
                return true;
            }
        }
        result = null;
        return false;
#endif
    }
    public static IPEndPoint AsIPEndPoint(this string s) => 
        s.TryParseAsIPEndPoint(out var endpoint) 
            ? endpoint 
            : throw new FormatException($"'{s}' is not a valid IP Endpoint");
   
}

My changes were to basically exchange Span<char> for string and make it an extension method of the class String itself. I've also conditionally compile to use Microsoft's implementation if it is available (NET Core 3.0 or greater).

The following nUnit tests show how to use the code:

[Test]
public void CanParseIpv4WithPort() {
    var sIp = "192.168.0.233:8080";
    if (sIp.TryParseAsIPEndPoint(out var endpoint)) {
        var expected = new IPEndPoint(new IPAddress(new byte[] { 192, 168, 0, 233 }), 8080);
        Assert.AreEqual(expected, endpoint);
    } else 
        Assert.Fail($"Failed to parse {sIp}");
}
[Test]
public void CanParseIpv6WithPort() {
    var sIp = "[2001:db8:85a3:8d3:1319:8a2e:370:7348]:443";
    if (sIp.TryParseAsIPEndPoint(out var endpoint)) {
        var expected = new IPEndPoint(IPAddress.Parse("2001:db8:85a3:8d3:1319:8a2e:370:7348"), 443);
        Assert.AreEqual(expected, endpoint);
    } else
        Assert.Fail($"Failed to parse {sIp}");
}

You can also use AsIpEndPoint which will throw an exception if it fails to parse the IP address and port (port is optional):

var ep = "127.0.0.1:9000".AsIPEndPoint();
Loudenvier
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