Using C++11 lambdas asynchronously, safely

Question

I\'ve come to C++11 from an Objective-C background, and one thing I\'m struggling to come to terms with is the different capturing semantics of C++11 lambdas vs Objective-C \"bl

Answer 1

Boost uses:

auto self(shared_from_this());
auto l = [this, self] { do(); };

Mentioned here: What's the reason of using auto self(shared_from_this()) variable in lambda function?

Answer 2

One of the founding principles of C++ is that you don't pay for what you don't use. That means in this case that contexts where taking a shared_ptr to this is unnecessary shouldn't incur any reference counting overhead. This also means that it shouldn't happen automatically even e.g. as a feature of enable_shared_from_this, since you might want to pass a short-lived lambda to an algorithm (for_each, etc.) in which case the lambda doesn't outlive its scope.

I'd suggest adapting the lambda-wrapper pattern; in that case it's used for move capture of a large object (How to capture std::unique_ptr "by move" for a lambda in std::for_each), but it can equally be used for shared capture of this:

template<typename T, typename F>
class shared_this_lambda {
  std::shared_ptr<T> t;  // just for lifetime
  F f;
public:
  shared_this_lambda(std::shared_ptr<T> t, F f): t(t), f(f) {}
  template<class... Args>
  auto operator()(Args &&...args)
  -> decltype(this->f(std::forward<Args>(args)...)) {
    return f(std::forward<Args>(args)...);
  }
};

template<typename T>
struct enable_shared_this_lambda {
  static_assert(std::is_base_of<std::enable_shared_from_this<T>, T>::value,
    "T must inherit enable_shared_from_this<T>");
  template<typename F>
  auto make_shared_this_lambda(F f) -> shared_this_lambda<T, F> {
    return shared_this_lambda<T, F>(
      static_cast<T *>(this)->shared_from_this(), f);
  }
  template<typename F>
  auto make_shared_this_lambda(F f) const -> shared_this_lambda<const T, F> {
    return shared_this_lambda<const T, F>(
      static_cast<const T *>(this)->shared_from_this(), f);
  }
};

Use by inheriting enable_shared_this_lambda in addition to enable_shared_from_this; you can then explicitly request that any long-lived lambdas take a shared this:

doSomethingAsynchronously(make_shared_this_lambda([this] {
  someMember_ = 42;
}));

Answer 3

Actually, there's one right answer to this problem. The answer has the exact same effect of binding with shared_from_this() (like when you do it with boost::asio::io_service). Think about it; what does binding with shared_from_this() do? It simple replaces this. So what prevents you from replacing this with shared_from_this() totally?

Following your example, which I updated to make the difference clearer, instead of this:

auto strongThis = shared_from_this();

doSomethingAsynchronously([strongThis, this] () {
  this->someMember_ = 42; //here, you're using `this`... that's wrong!
});

Do this:

auto strongThis = shared_from_this();

doSomethingAsynchronously([strongThis] () //notice, you're not passing `this`!
{
  strongThis->someMember_ = 42;            
});

The only cost here is that you're gonna have to prefix everything with strongThis->. But this is the most meaningful way to do it.