如何将 C++ lambda 传递给需要函数指针和上下文的 C 回调？

Question

I'm trying to register a callback in a C-API that uses the standard function-pointer+context paradigm. Here's what the api looks like:

void register_callback(void(*callback)(void *), void * context);

What I'd really like to do is be able to register a C++ lambda as the callback. Additionally, I want the lambda to be one that has captured variables (ie. can't be converted to a straight stateless std::function )

What kind of adapter code would I need to write to be able to register a lambda as the callback?

Answer 1

The simple approach is to stick the lambda into a std::function<void()> which is kept somewhere. Potentially it is allocated on the heap and merely referenced by the void* registered with the entity taking the callback. The callback would then simply be a function like this:

extern "C" void invoke_function(void* ptr) {
    (*static_cast<std::function<void()>*>(ptr))();
}

Note that std::function<S> can hold function objects with state, eg, lambda functions with a non-empty capture. You could register a callback like this:

register_callback(&invoke_function,
  new std::function<void()>([=](){ ... }));

Answer 2

The most efficient way is to voidify the lambda directly.

#include <iostream>
#include <tuple>
#include <memory>

template<class...Args>
struct callback {
  void(*function)(void*, Args...)=nullptr;
  std::unique_ptr<void, void(*)(void*)> state;
};
template<typename... Args, typename Lambda>
callback<Args...> voidify( Lambda&& l ) {
  using Func = typename std::decay<Lambda>::type;
  std::unique_ptr<void, void(*)(void*)> data(
    new Func(std::forward<Lambda>(l)),
    +[](void* ptr){ delete (Func*)ptr; }
  );
  return {
    +[](void* v, Args... args)->void {
      Func* f = static_cast< Func* >(v);
      (*f)(std::forward<Args>(args)...);
    },
    std::move(data)
  };
}

void register_callback( void(*function)(void*), void * p ) {
  function(p); // to test
}
void test() {
  int x = 0;
  auto closure = [&]()->void { ++x; };
  auto voidified = voidify(closure);
  register_callback( voidified.function, voidified.state.get() );
  register_callback( voidified.function, voidified.state.get() );
  std::cout << x << "\n";
}
int main() {
  test();
}

here voidify takes a lambda and (optionally) a list of arguments, and generates a traditional C-style callback- void* pair. The void* is owned by a unique_ptr with a special deleter so its resources are properly cleaned up.

The advantage of this over a std::function solution is efficiency -- I eliminated one level of run-time indirection. The lifetime that the callback is valid is also clear, in that it is in the std::unique_ptr<void, void(*)(void*)> returned by voidify .

unique_ptr<T,D> s can be move d into shared_ptr<T> if you want a more complex lifetime.

---

The above mixes lifetime with data, and type erasure with utility. We can split it:

template<class Lambda, class...Args>
struct callback {
  void(*function)(void*, Args...)=nullptr;
  Lambda state;
};

template<typename... Args, typename Lambda>
callback<typename std::decay<Lambda>::type, Args...> voidify( Lambda&& l ) {
  using Func = typename std::decay<Lambda>::type;
  return {
    +[](void* v, Args... args)->void {
      Func* f = static_cast< Func* >(v);
      (*f)(std::forward<Args>(args)...);
    },
    std::forward<Lambda>(l)
  };
}

Now voidify does not allocate. Simply store your voidify for the lifetime of the callback, passing a pointer-to- second as your void* along side the first function pointer.

If you need to store this construct off the stack, converting the lambda to a std::function may help. Or use the first variant above.

void test() {
  int x = 0;
  auto closure = [&]()->void { ++x; };
  auto voidified = voidify(closure);
  register_callback( voidified.function, &voidified.state );
  register_callback( voidified.function, &voidified.state );
  std::cout << x << "\n";
}

Answer 3

A lambda function is compatible with C-callback function as long as it doesn't have capture variables.
Force to put something new to old one with new way doesn't make sense.
How about following old-fashioned way?

typedef struct
{
  int cap_num;
} Context_t;

int cap_num = 7;

Context_t* param = new Context_t;
param->cap_num = cap_num;   // pass capture variable
register_callback([](void* context) -> void {
    Context_t* param = (Context_t*)context;
    std::cout << "cap_num=" << param->cap_num << std::endl;
}, param);

Answer 4

A very simple way to get use a lamda function as a C funtion pointer is this:

auto fun=+[](){ //add code here }

see this answer for a example. https://stackoverflow.com/a/56145419/513481