I'm looking at the source code for a server using SocketAsyncEventArgs, and I'm trying to figure out how this wouldn't cause a stack overflow:
So this code is called to allow the socket to accept an incoming connection (scroll down to the bottom to see what I mean):
/// <summary>
/// Begins an operation to accept a connection request from the client.
/// </summary>
/// <param name="acceptEventArg">The context object to use when issuing
/// the accept operation on the server's listening socket.</param>
private void StartAccept(SocketAsyncEventArgs acceptEventArg)
{
if (acceptEventArg == null)
{
acceptEventArg = new SocketAsyncEventArgs();
acceptEventArg.Completed += new EventHandler<SocketAsyncEventArgs>(OnAcceptCompleted);
}
else
{
// Socket must be cleared since the context object is being reused.
acceptEventArg.AcceptSocket = null;
}
this.semaphoreAcceptedClients.WaitOne();
Boolean willRaiseEvent = this.listenSocket.AcceptAsync(acceptEventArg);
if (!willRaiseEvent)
{
this.ProcessAccept(acceptEventArg);
}
}
Then this code gets called once a connection is actually accepted (see last line):
/// <summary>
/// Process the accept for the socket listener.
/// </summary>
/// <param name="e">SocketAsyncEventArg associated with the completed accept operation.</param>
private void ProcessAccept(SocketAsyncEventArgs e)
{
if (e.BytesTransferred > 0)
{
Interlocked.Increment(ref this.numConnectedSockets);
Console.WriteLine("Client connection accepted. There are {0} clients connected to the server",
this.numConnectedSockets);
}
// Get the socket for the accepted client connection and put it into the
// ReadEventArg object user token.
SocketAsyncEventArgs readEventArgs = this.readWritePool.Pop();
readEventArgs.UserToken = e.AcceptSocket;
// As soon as the client is connected, post a receive to the connection.
Boolean willRaiseEvent = e.AcceptSocket.ReceiveAsync(readEventArgs);
if (!willRaiseEvent)
{
this.ProcessReceive(readEventArgs);
}
// Accept the next connection request.
this.StartAccept(e); // <==== tail end recursive?
}
Look at the last line. It calls the top function again. How does this not overflow the stack by ping-ponging back and forth between these 2 functions? It seems to be tail end recursion, but this isn't Haskell so I don't see how this would work.
It was my understanding that these weren't fired in threads but where just executed one at a time by the cpu.