SslStream.SslProtocol always returns TLS (TLS 1.0)

Question

Trying to retrieve encryption connection details from website. Below is the code.

string strWebsiteName = "social.msdn.microsoft.com"
TcpClient _myClient = new TcpClient();
SslStream _myStream;
_myClient.Connect(strWebsiteName, 443);
 _myStream = new SslStream(_myClient.GetStream());
_myStream.AuthenticateAsClient(strWebsiteName);

Console.WriteLine("SSL Protocol : " + _myStream.SslProtocol.ToString().ToUpper());

Though social.msdn.microsoft.com uses TLS 1.2, the code returns TLS which is TLS 1.0. Can anyone help where I am missing.

Answer 1

Hi Jagadheesh Venkatesan,

>> Though social.msdn.microsoft.com uses TLS 1.2, the code returns TLS which is TLS 1.0. Can anyone help where I am missing.

According to the following link that’s about the Remarks for SslStream.AuthenticateAsClient Method (String) you could know that this method use Default SslProtocol to authenticate. So the SslProtocol would always be Transport Layer Security (TLS) 1.0 if your client has enable it. About why it would not be Secure Sockets Layer (SSL) 3.0, you should know that the SSL server would return the highest version supported by both peers.

https://msdn.microsoft.com/en-us/library/ms145060(v=vs.110).aspx#Anchor_2

Best Regards,

Albert Zhang

Answer 2

Btw, althougth not specified in your code, just want to point out that if you want to run it on a system without .NET 4.5+ version of runtime installed, TLS1.0 is the best you can get (TLS1.1 and TLS1.2 support is added in .NET v4.5).

Answer 3

Thank you Albert. I was able to get an answer. Below is the snippet. We can include desired protocol version and it returns the highest version supported. 

_myStream.AuthenticateAsClient(strWebsiteName, new X509CertificateCollection(),SslProtocols.Tls12 | SslProtocols.Tls11 |SslProtocols.Default,false);

Answer 4

Yes Cheong. Working on .NET 4.5.1

Answer 5

Hi Jagadheesh Venkatesan,

I am glad to know you solved this problem and thanks for sharing the solution. 

It will be very beneficial for other community members who have the similar questions.

Best Regards,

Albert Zhang

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