Convert C++ function pointer to c function pointer

Question

I am developing a C++ application using a C library. I have to send a pointer to function to the C library.

This is my class:

 class MainWindow : pub         


        

Answer 1

@Snps answer is perfect! But as @DXM mentioned it can hold only one callback. I've improved it a little, now it can keep many callbacks of the same type. It's a little bit strange, but works perfect:

 #include <type_traits>

template<typename T>
struct ActualType {
    typedef T type;
};
template<typename T>
struct ActualType<T*> {
    typedef typename ActualType<T>::type type;
};

template<typename T, unsigned int n,typename CallerType>
struct Callback;

template<typename Ret, typename ... Params, unsigned int n,typename CallerType>
struct Callback<Ret(Params...), n,CallerType> {
    typedef Ret (*ret_cb)(Params...);
    template<typename ... Args>
    static Ret callback(Args ... args) {
        func(args...);
    }

    static ret_cb getCallback(std::function<Ret(Params...)> fn) {
        func = fn;
        return static_cast<ret_cb>(Callback<Ret(Params...), n,CallerType>::callback);
    }

    static std::function<Ret(Params...)> func;

};

template<typename Ret, typename ... Params, unsigned int n,typename CallerType>
std::function<Ret(Params...)> Callback<Ret(Params...), n,CallerType>::func;

#define GETCB(ptrtype,callertype) Callback<ActualType<ptrtype>::type,__COUNTER__,callertype>::getCallback

Now you can just do something like this:

typedef void (cb_type)(uint8_t, uint8_t);
class testfunc {
public:
    void test(int x) {
        std::cout << "in testfunc.test " <<x<< std::endl;
    }

    void test1(int x) {
        std::cout << "in testfunc.test1 " <<x<< std::endl;
    }

};

cb_type* f = GETCB(cb_type, testfunc)(std::bind(&testfunc::test, tf, std::placeholders::_2));

cb_type* f1 = GETCB(cb_type, testfunc)(
                std::bind(&testfunc::test1, tf, std::placeholders::_2));


f(5, 4);
f1(5, 7);

Answer 2

If I recall it correctly, Only static methods of a class can be accessed via "normal" C pointer to function syntax. So try to make it static. The pointer to a method of a class needs extra information, such as the "object" (this) which has no meaning for a pure C method.

The FAQ shown here has good explanation and a possible (ugly) solution for your problem.

Answer 3

You can't pass a function pointer to a non-static member function. What you can do is to create a static or global function that makes the call with an instance parameter.

Here's an example I find useful which uses a helper class with two members: a function wrapper and a callback function that calls the wrapper.

template <typename T>
struct Callback;

template <typename Ret, typename... Params>
struct Callback<Ret(Params...)> {
    template <typename... Args>
    static Ret callback(Args... args) { return func(args...); }
    static std::function<Ret(Params...)> func;
};

// Initialize the static member.
template <typename Ret, typename... Params>
std::function<Ret(Params...)> Callback<Ret(Params...)>::func;

Using this you can store any callable, even non-static member functions (using std::bind) and convert to a c-pointer using the Callback::callback function. E.g:

struct Foo {
    void print(int* x) { // Some member function.
        std::cout << *x << std::endl;
    }
};

int main() {
    Foo foo; // Create instance of Foo.

    // Store member function and the instance using std::bind.
    Callback<void(int*)>::func = std::bind(&Foo::print, foo, std::placeholders::_1);

    // Convert callback-function to c-pointer.
    void (*c_func)(int*) = static_cast<decltype(c_func)>(Callback<void(int*)>::callback);

    // Use in any way you wish.
    std::unique_ptr<int> iptr{new int(5)};
    c_func(iptr.get());
}

Answer 4

I've got an idea (not entirely standard-compliant, as extern "C" is missing):

class MainWindow;

static MainWindow* instance;

class MainWindow
{
public:
  MainWindow()
  {
    instance = this;

    registerCallback([](int* arg){instance->...});
  }
};

You will have problems if multiple instances of MainWindow are instantiated.

Answer 5

@Snps answer is great. I extended it with a maker function that creates a callback, as I always use void callbacks without parameters:

typedef void (*voidCCallback)();
template<typename T>
voidCCallback makeCCallback(void (T::*method)(),T* r){
  Callback<void()>::func = std::bind(method, r);
  void (*c_function_pointer)() = static_cast<decltype(c_function_pointer)>(Callback<void()>::callback);
  return c_function_pointer;
}

From then on, you can create your plain C callback from within the class or anywhere else and have the member called:

voidCCallback callback = makeCCallback(&Foo::print, this);
plainOldCFunction(callback);

Answer 6

The short answer is: you can convert a member function pointer to an ordinary C function pointer using std::mem_fn.

That is the answer to the question as given, but this question seems to have a confused premise, as the asker expects C code to be able to call an instance method of MainWindow without having a MainWindow*, which is simply impossible.

If you use mem_fn to cast MainWindow::on_btn_clicked to a C function pointer, then you still a function that takes a MainWindow* as its first argument.

void (*window_callback)(MainWindow*,int*) = std::mem_fn(&MainWindow::on_btn_clicked);

That is the answer to the question as given, but it doesn't match the interface. You would have to write a C function to wrap the call to a specific instance (after all, your C API code knows nothing about MainWindow or any specific instance of it):

void window_button_click_wrapper(int* arg)
{
    MainWindow::inst()->on_btn_clicked(arg);
}

This is considered an OO anti-pattern, but since the C API knows nothing about your object, it's the only way.