How can I use lambda function within itself?

Question

I have this code and don't know if what I would like to achieve is possible.

_acceptor.async_accept(
    _connections.back()->socket(),
    [this](const boost::system::error_code& ec)
    {
        _connections.push_back(std::make_shared<TcpConnection>(_acceptor.get_io_service()));
        _acceptor.async_accept(_connections.back()->socket(), this_lambda_function);
    }
);

Once a socket is accepted, I would like to reuse the handler (aka the lambda function). Is this possible? Is there a better way to accomplish this?

+1 Very interesting question. I hadn't thought of that before. — templatetypedef
– templatetypedef, Commented Apr 8, 2012 at 19:21
groups.google.com/group/comp.lang.c++.moderated/browse_thread/… — Anonymous
– Anonymous, Commented Apr 8, 2012 at 19:27
Not related to your question, but you should know that leading underscores (and two adjacent underscores) are reserved and shouldn't be used for application identifiers. — Marc
– Marc, Commented Apr 10, 2012 at 23:44

Nicol Bolas · Accepted Answer · 2012-04-08 19:35:17Z

9

You have to store a copy of the lambda in itself, using std::function<> (or something similar) as an intermediary:

std::function<void(const boost::system::error_code&)> func;
func = [&func, this](const boost::system::error_code& ec)
{
    _connections.push_back(std::make_shared<TcpConnection>(_acceptor.get_io_service()));
    _acceptor.async_accept(_connections.back()->socket(), func);
}

_acceptor.async_accept(_connections.back()->socket(), func);

But you can only do it by reference; if you try to capture it by value, it won't work. This means you have to limit the usage of such a lambda to uses were capture-by-reference will make sense. So if you leave this scope before your async function is finished, it'll break.

Your other alternative is to create a proper functor rather than a lambda. Ultimately, lambdas can't do everything.

answered Apr 8, 2012 at 19:35

Nicol Bolas

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1 Comment

Nicol Bolas Over a year ago

@balki: No. It is legal in C/C++ to initialize a variable with an expression that uses the variable name. However, this is shut off when dealing with auto variables since the variable name doesn't have a type until the expression's type can be determined.

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