Get function pointer from std::function when using std::bind

Question

I'm trying to use std::function in conjunction with std::bind, but I'm having some problems.

This works:

#include <functional>
#include <iostream>

void print() {
    std::cout << 2;
}

int main() {
    std::function<void ()> foo = print;
    (*foo.target<void (*)()>())(); //prints 3
}

This crashes at the second line of main:

#include <functional>
#include <iostream>

void print (int i) {
    std::cout << i;
}

int main() {
    std::function<void ()> foo = std::bind (print, 2);
    (*foo.target<void (*)()>())();
}

I'm really holding the std::function<void ()> and need to be able to return the function; not just call it. I expect the usage would be something like this:

#include <functional>
#include <iostream>

void print (int i) {
    std::cout << i;
}

int main() {
    Container c (std::bind (print, 2));

    //I would expect the original
    c.func() (3); //prints 3

    if (c.func() == print) /* this is what I'm mostly getting at */
}

Is there any way to get the original function to return it, or an alternative? It does kind of conflict with the return type as well, as void (*)() matches the bound signature quite nicely.

Answer 1

This is quite impossible. The whole reason that std::function exists is that function pointers suck horrifically and should never, ever, be used by anyone, ever again, except for the doomed souls bearing the Burning Standards of Hell C interoperation, because they cannot handle functions with state.

A std::function<void()> cannot, in the general case, be converted to a void(*)(). The only reason this works in the first example is because it happens to be a void(*)() originally.

Answer 2

This can be achieved using a little template meta-programming. I recently had use for this while writing a generic C++ wrapper around OpenGL GLUT (which depends on callback function pointers). The approach:

1. Instantiate an instance of a singleton template type.
2. Store your std::function as a member of to the singleton instance
3. Invoke your std::function through a static member function (static member functions and free functions have the same type, so the "invoke" function can be used as a free function pointer)

Tested under C++11 on GCC 4.8.

#include <unistd.h>
#include <thread>
#include <chrono>
#include <mutex>
#include <functional>
#include <iostream>
#include <cmath>

template <const size_t _UniqueId, typename _Res, typename... _ArgTypes>
struct fun_ptr_helper
{
public:
    typedef std::function<_Res(_ArgTypes...)> function_type;

    static void bind(function_type&& f)
    { instance().fn_.swap(f); }

    static void bind(const function_type& f)
    { instance().fn_=f; }

    static _Res invoke(_ArgTypes... args)
    { return instance().fn_(args...); }

    typedef decltype(&fun_ptr_helper::invoke) pointer_type;
    static pointer_type ptr()
    { return &invoke; }

private:
    static fun_ptr_helper& instance()
    {
        static fun_ptr_helper inst_;
        return inst_;
    }

    fun_ptr_helper() {}

    function_type fn_;
};

template <const size_t _UniqueId, typename _Res, typename... _ArgTypes>
typename fun_ptr_helper<_UniqueId, _Res, _ArgTypes...>::pointer_type
get_fn_ptr(const std::function<_Res(_ArgTypes...)>& f)
{
    fun_ptr_helper<_UniqueId, _Res, _ArgTypes...>::bind(f);
    return fun_ptr_helper<_UniqueId, _Res, _ArgTypes...>::ptr();
}

template<typename T>
std::function<typename std::enable_if<std::is_function<T>::value, T>::type>
make_function(T *t)
{
    return {t};
}

int main()
{
    std::cout << (void*)get_fn_ptr<0>(make_function(::sin))<<std::endl;
    return 0;
}

Answer 3

You can't get a function pointer out of an std::function, as there may not even be one. It could be a member function pointer instead, or an object that implements operator().