store a lambda that captures this - c++17

Question

Using C++ 17, I'm looking for a way to store a lambda that captures the this pointer, without using std::function<>. The reason to not using std::function<> is that I need the guaranty that no dynamic memory allocations are used. The purpose of this, is to be able to define some asynchronous program flow. Example:
class foo {
public:
void start() {
timer(1ms, [this](){
set_pin(1,2);
timer(1ms, [this](){
set_pin(2,1);
}
}
}
private:
template < class Timeout, class Callback >
void timer( Timeout to, Callback&& cb ) {
cb_ = cb;
// setup timer and call cb_ one timeout reached
...
}
??? cb_;
};
Edit: Maybe it's not really clear: std::function<void()> would do the job, but I need / like to have the guaranty, that no dynamic allocations happens as the project is in the embedded field. In practice std::function<void()> seems to not require dynamic memory allocation, if the lambda just captures this. I guess this is due to some small object optimizations, but I would like to not rely on that.

Answer 1

You can write your own function_lite to store the lambda, then you can use static_assert to check the size and alignment requirements are satisfied:
#include <cstddef>
#include <new>
#include <type_traits>
class function_lite {
static constexpr unsigned buffer_size = 16;
using trampoline_type = void (function_lite::*)() const;
trampoline_type trampoline;
trampoline_type cleanup;
alignas(std::max_align_t) char buffer[buffer_size];
template <typename T>
void trampoline_func() const {
auto const obj =
std::launder(static_cast<const T*>(static_cast<const void*>(buffer)));
(*obj)();
}
template <typename T>
void cleanup_func() const {
auto const obj =
std::launder(static_cast<const T*>(static_cast<const void*>(buffer)));
obj->~T();
}
public:
template <typename T>
function_lite(T t)
: trampoline(&function_lite::trampoline_func<T>),
cleanup(&function_lite::cleanup_func<T>) {
static_assert(sizeof(T) <= buffer_size);
static_assert(alignof(T) <= alignof(std::max_align_t));
new (static_cast<void*>(buffer)) T(t);
}
~function_lite() { (this->*cleanup)(); }
function_lite(function_lite const&) = delete;
function_lite& operator=(function_lite const&) = delete;
void operator()() const { (this->*trampoline)(); }
};
int main() {
int x = 0;
function_lite f([x] {});
}
Note: this is not copyable; to add copy or move semantics you will need to add new members like trampoline and cleanup which can properly copy the stored object.

Answer 2

There is no drop in replacement in the language or the standard library.
Every lambda is a unique type in the typesystem. Technically you may have a lambda as a member, but then its type is fixed. You may not assign other lambdas to it.
If you really want to have an owning function wrapper like std::function, you need to write your own. Actually you want a std::function with a big enough small-buffer-optimization buffer.
Another approach would be to omit the this capture and pass it to the function when doing the call. So you have a captureless lambda, which is convertible to a function pointer which you can easily store. I would take this route and adapter complexer ways if really nessessary.
it would look like this (i trimmed down the code a bit):
class foo
{
public:
void start()
{
timer(1, [](foo* instance)
{
instance->set_pin(1,2);
});
}
private:
template < class Timeout, class Callback >
void timer( Timeout to, Callback&& cb )
{
cb_ = cb;
cb_(this); // call the callback like this
}
void set_pin(int, int)
{
std::cout << "pin set\n";
}
void(*cb_)(foo*);
};