Understand the usage of timeout in beast::tcp_stream? - c++

Question

Reference:
https://www.boost.org/doc/libs/1_78_0/libs/beast/example/websocket/client/async/websocket_client_async.cpp
https://www.boost.org/doc/libs/1_78_0/libs/beast/doc/html/beast/using_io/timeouts.html
https://www.boost.org/doc/libs/1_78_0/libs/beast/doc/html/beast/ref/boost__beast__tcp_stream.html
void on_resolve(beast::error_code ec, tcp::resolver::results_type results)
{
if(ec) return fail(ec, "resolve");
// Set the timeout for the operation
beast::get_lowest_layer(ws_).expires_after(std::chrono::seconds(30));
// Make the connection on the IP address we get from a lookup
beast::get_lowest_layer(ws_).async_connect(
results, beast::bind_front_handler(
&session::on_connect, shared_from_this()));
}
void on_connect(beast::error_code ec, tcp::resolver::results_type::endpoint_type ep)
{
if(ec) return fail(ec, "connect");
// Turn off the timeout on the tcp_stream, because
// the websocket stream has its own timeout system.
// beast::get_lowest_layer(ws_).expires_never(); // Note: do NOT call this line for this question!!!
...
host_ += ':' + std::to_string(ep.port());
// Perform the websocket handshake
ws_.async_handshake(host_, "/",
beast::bind_front_handler(&session::on_handshake, shared_from_this()));
}
Question 1>
Will the timeout of beast::tcp_stream continue to work after a previous asynchronous operation finishes on time?
For example,
In above example, the timeout will expire after 30 seconds. If async_connect doesn't finish within 30 seconds, session::on_connect will receive an error::timeout as the value of ec. Let's assume the async_connect takes 10 seconds,
can I assume that async_handshake needs to finish within 20(i.e. 30-10) seconds otherwise a error::timeout will be sent to session::on_handshake? I infer to this idea based on the comments within on_connect function(i.e.
Turn off the timeout on the tcp_stream
). In other words, a timeout will only be turned off after it finishes the specified expiration period or is disabled by expires_never. Is my understanding correct?
Question 2> Also I want to know what a good pattern I should use for timeout in both async_calling and async_callback functions.
When we call an async_calling operation:
void func_async_calling()
{
// set some timeout here(i.e. XXXX seconds)
Step 1> beast::get_lowest_layer(ws_).expires_after(std::chrono::seconds(XXXX));
Step 2> ws_.async_operation(..., func_async_callback, )
Step 3> beast::get_lowest_layer(ws_).expires_never();
}
When we define a async_callback handle for an asynchronous operation:
void func_async_callback()
{
Step 1>Either call
// Disable the timeout for the next logical operation.
beast::get_lowest_layer(ws_).expires_never();
or
// Enable a new timeout
beast::get_lowest_layer(ws_).expires_after(std::chrono::seconds(YYYY));
Step 2> call another asynchronous function
Step 3> beast::get_lowest_layer(ws_).expires_never();
}
Does this make sense?
Thank you

Answer 1

Question 1
Yes that's correct. The linked page has the confirmation:
// The timer is still running. If we don't want the next
// operation to time out 30 seconds relative to the previous
// call to `expires_after`, we need to turn it off before
// starting another asynchronous operation.
stream.expires_never();
Question 2
That looks fine. The only subtleties I can think of are
often, because of Thread Safety often the initiation as well as the completion happen on the same (implicit) strand.
If that's the case, then in your completion handler example, the expires_never(); would be redundant.
If the completion handler is not on the same strand, you want to actively avoid touching the expiry, because that would be a data race
An alternative pattern is to set the expiry only once for a lengthier episode (e.g. an multi-message conversation between client/server). Obviously in this pattern, nobody would touch the expiry after initial setting. This seems pretty obvious, but I thought I'd mention it before someone casts this pattern in stone to never think about it again.
Always do what you need, prefer simple code. I think your basic understanding of the feature is right. (No wonder, this documentation is a piece of art).