WCF error starting as Windows service

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I am trying to start a WCF as Windows service, but it returns me the following error:

HTTP could not register the URL http://+:80/. Your process has no access rights to this namespace (see website http://go.microsoft.com/fwlink/? Linkid=70353 for details).

How can I solve this problem, when it was a common executable, it worked perfectly, now as I am putting it as windows service, it keeps returning this error.

I’m doing it this way to start:

ServiceHost host = new WebServiceHost(typeof(Server));
            WebHttpBinding binding = new WebHttpBinding();
            binding.MaxReceivedMessageSize = 999999999;
            ServiceEndpoint point = host.AddServiceEndpoint(typeof(IServer), binding, url);
            point.Behaviors.Add(new WebHttpBehavior());
            host.Open();
            btnSalvar.Log(new string[1] { "CONECTADO" });

The problem is not the code, because as I already informed, it works on a common exe, the error happens after putting as windows service.

c# wcf

  • Is the Logon account in the service an account with permissions? You tried with an administrator account only to validate if the service starts?

  • Yes the account is an administrator ..

  • Then it must be a problem to reserve the url+namepace+port. In the message link you have the commands to grant permission to the user who is running the service. For example in Windows Server 2008 forward or Windows 7 forward, the command is this: netsh http add urlacl url=http://+:80/MyUri user=DOMAIN\user just put the login dominance you are using in the service

  • @Ricardopunctual I did so, but continues with the same mistake.

1 answer

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I can not find the link where I found this answer, but this is how I solved the problem:

I was working on a WFC test service hosted on Windows Service (managed). So I added another base address for HTTP. When I started the service again, I received the same error. Then I changed the service account from "Local Service" to "Local System" and this solved the problem.

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