boost::lockfree::queue of functions? - c++

Question

I'd like to build a simple boost::lockfree::queue of functions that take no arguments and return no values.
It appears that boost::lockfree::queue requires the item type to be trivially assignable and destructible, requirements that boost::function<void ()> unfortunately doesn' meet.
In the spirit of https://stackoverflow.com/a/21406186/393756, I'm now trying to achieve this by a boost::lockfree::queue of plain function pointers:
boost::lockfree::queue<void (*)()> queue;
Can I push a boost::function<void ()> into this queue? If so, how?

Answer 1

Can I push a boost::function<void()> into this queue?
Not directly, as boost::function<void()> is an heavyweight owning type-erased wrapper that is not implicitly convertible to a function pointer and also stores some data.
If you need a trivially-assignable an trivially-destructible type that can refer to any function object, you could implement a function_view class that points to some function object without owning it. If you're careful with lifetimes and guarantee that function_view always points to "live objects" you can safely store instances of that in your queue.
Conceptually, function_view is a pair of pointers. I have an implementation in my "passing functions to functions" article, which I'm pasting below:
template <typename TReturn, typename... TArgs>
class function_view<TReturn(TArgs...)> final
{
private:
using signature_type = TReturn(void*, TArgs...);
void* _ptr;
TReturn (*_erased_fn)(void*, TArgs...);
public:
template <typename T, typename = std::enable_if_t<
std::is_callable<T&(TArgs...)>{} &&
!std::is_same<std::decay_t<T>, function_view>{}>>
function_view(T&& x) noexcept : _ptr{(void*)std::addressof(x)}
{
_erased_fn = [](void* ptr, TArgs... xs) -> TReturn {
return (*reinterpret_cast<std::add_pointer_t<T>>(ptr))(
std::forward<TArgs>(xs)...);
};
}
decltype(auto) operator()(TArgs... xs) const
noexcept(noexcept(_erased_fn(_ptr, std::forward<TArgs>(xs)...)))
{
return _erased_fn(_ptr, std::forward<TArgs>(xs)...);
}
};
This class passes the following tests:
using type = function_view<void()>;
static_assert(is_trivially_assignable<type, type>{});
static_assert(is_trivially_destructible<type>{});
live example on wandbox

Answer 2

no, but you can use dynamic memory allocation + type erasure for that cause:
struct callback_back{
virtual void execute() = 0;
~callback_base() = default;
};
template<class F>
class callback{
private:
F m_function;
public:
callback(F&& function) : m_function(std::forward<F>(function)){}
virtual void execute() {
m_function();
}
}
template<class F>
std::unique_ptr<callback_base> make_callback(F&& f){
return std::unique_ptr<callback_base>(
new callback<F>(std::forward<F>(f));
);
}
use callback_base as a noexcept-movable type (aka boost::lockfree::queue<std::unique_ptr<callback_base>>).

Answer 3

The only way I found so far is to make a raw pointer of the function object
boost::lockfree::queue<std::function<void(void)> *> tasks_; // the queue
// let f = stack allocated std::function<T(T)> instance
tasks_.push(new std::function<void(void)>(f));
// pop
std::function<void(void)> * f;
tasks_.pop(f);
// execute in try/catch make sure to delete in case of exception?
(*f)();
// you should delete f here if you are done with it
// in the destructor of the class that owns tasks_ you should delete the remaining std::function instances
The challenge here is to when to delete this instance with exception safety in mind