Question

I am using the C++ boost asio library, where I listen to new connections on the socket. On getting a connection I process the request and then listen for a new connection on another socket in a loop.

while (true)
{
    tcp::socket soc(this->blitzIOService);
    this->blitzAcceptor.listen();
    boost::system::error_code ec;
    this->blitzAcceptor.accept(soc,ec);
    if (ec)
    {
        // Some error occured
        cerr << "Error Value: " << ec.value() << endl;
        cerr << "Error Message: " << ec.message() << endl;
        soc.close();
        break;
    }
    else
    {
        this->HandleRequest(soc);
        soc.shutdown(tcp::socket::shutdown_both);
        soc.close();
    }
}

According to my understanding it should always block at this->blitzAcceptor.accept(soc,ec); and everytime a new connection is made it should handle it in this->HandleRequest(soc); and again block at this->blitzAcceptor.accept(soc,ec);

But what I see is this that for the first time it will block at this->blitzAcceptor.accept(soc,ec) and when a new connection is made it will handle the request, but instead of blocking again at this->blitzAcceptor.accept(soc,ec) it will go ahead into this->HandleRequest(soc); and block at soc.receive(); inside.

This doesn’t happen always, but happens most of the time. What could be the reason to this behavior, and how can I ensure that it always block at this->blitzAcceptor.accept(soc,ec) until a new request is made?

Answer 1

What could be the reason to this
behavior?

This behavior is entirely dependent on the client code. If it connects, but does not send a request, the server with block when receiving data.

how can I ensure that it always block
at this->blitzAcceptor.accept(soc,ec)
until a new request is made?

You can’t. But your server can initiate a timeout that starts immediately after accepting the connection. If the client does not send a request within that duration, close the socket. To do that, you should switch to using asynchronous methods rather than synchronous methods.