Passing capturing lambda as function pointer

Shafik Yaghmour's answer correctly explains why the lambda cannot be passed as a function pointer if it has a capture. I'd like to show two simple fixes for the problem.

1. Use std::function instead of raw function pointers.

   This is a very clean solution. Note however that it includes some additional overhead for the type erasure (probably a virtual function call).

   #include <functional>
   #include <utility>
   
   struct Decide
   {
     using DecisionFn = std::function<bool()>;
     Decide(DecisionFn dec) : dec_ {std::move(dec)} {}
     DecisionFn dec_;
   };
   
   int
   main()
   {
     int x = 5;
     Decide greaterThanThree { [x](){ return x > 3; } };
   }
   

2. Use a lambda expression that doesn't capture anything.

   Since your predicate is really just a boolean constant, the following would quickly work around the current issue. See this answer for a good explanation why and how this is working.

   // Your 'Decide' class as in your post.
   
   int
   main()
   {
     int x = 5;
     Decide greaterThanThree {
       (x > 3) ? [](){ return true; } : [](){ return false; }
     };
   }
   

Lambda expressions, even captured ones, can be handled as a function pointer (pointer to member function).

It is tricky because an lambda expression is not a simple function. It is actually an object with an operator().

When you are creative, you can use this! Think of an "function" class in style of std::function. If you save the object you also can use the function pointer.

To use the function pointer, you can use the following:

int first = 5;
auto lambda = [=](int x, int z) {
    return x + z + first;
};
int(decltype(lambda)::*ptr)(int, int)const = &decltype(lambda)::operator();
std::cout << "test = " << (lambda.*ptr)(2, 3) << std::endl;

To build a class that can start working like a "std::function", first you need a class/struct than can store object and function pointer. Also you need an operator() to execute it:

// OT => Object Type
// RT => Return Type
// A ... => Arguments
template<typename OT, typename RT, typename ... A>
struct lambda_expression {
    OT _object;
    RT(OT::*_function)(A...)const;

    lambda_expression(const OT & object)
        : _object(object), _function(&decltype(_object)::operator()) {}

    RT operator() (A ... args) const {
        return (_object.*_function)(args...);
    }
};

With this you can now run captured, non-captured lambdas, just like you are using the original:

auto capture_lambda() {
    int first = 5;
    auto lambda = [=](int x, int z) {
        return x + z + first;
    };
    return lambda_expression<decltype(lambda), int, int, int>(lambda);
}

auto noncapture_lambda() {
    auto lambda = [](int x, int z) {
        return x + z;
    };
    return lambda_expression<decltype(lambda), int, int, int>(lambda);
}

void refcapture_lambda() {
    int test;
    auto lambda = [&](int x, int z) {
        test = x + z;
    };
    lambda_expression<decltype(lambda), void, int, int>f(lambda);
    f(2, 3);

    std::cout << "test value = " << test << std::endl;
}

int main(int argc, char **argv) {
    auto f_capture = capture_lambda();
    auto f_noncapture = noncapture_lambda();

    std::cout << "main test = " << f_capture(2, 3) << std::endl;
    std::cout << "main test = " << f_noncapture(2, 3) << std::endl;

    refcapture_lambda();

    system("PAUSE");
    return 0;
}

This code works with VS2015

Update 04.07.17:

template <typename CT, typename ... A> struct function
: public function<decltype(&CT::operator())(A...)> {};

template <typename C> struct function<C> {
private:
    C mObject;

public:
    function(const C & obj)
        : mObject(obj) {}

    template<typename... Args> typename 
    std::result_of<C(Args...)>::type operator()(Args... a) {
        return this->mObject.operator()(a...);
    }

    template<typename... Args> typename 
    std::result_of<const C(Args...)>::type operator()(Args... a) const {
        return this->mObject.operator()(a...);
    }
};

namespace make {
    template<typename C> auto function(const C & obj) {
        return ::function<C>(obj);
    }
}

int main(int argc, char ** argv) {
   auto func = make::function([](int y, int x) { return x*y; });
   std::cout << func(2, 4) << std::endl;
   system("PAUSE");
   return 0;
}

Capturing lambdas cannot be converted to function pointers, as this answer pointed out.

However, it is often quite a pain to supply a function pointer to an API that only accepts one. The most often cited method to do so is to provide a function and call a static object with it.

static Callable callable;
static bool wrapper()
{
    return callable();
}

This is tedious. We take this idea further and automate the process of creating wrapper and make life much easier.

#include<type_traits>
#include<utility>
#include<new>

template<int, typename Callable, typename Ret, typename... Args>
auto fnptr_(Callable&& c, Ret (*)(Args...))
{
    static std::decay_t<Callable> storage = std::forward<Callable>(c);
    static bool used = false;
    if(used)
    {
        using type = decltype(storage);
        storage.~type();
        new (&storage) type(std::forward<Callable>(c));
    }
    used = true;

    return [](Args... args) -> Ret {
        auto& c = *std::launder(&storage);
        return Ret(c(std::forward<Args>(args)...));
    };
}

template<typename Fn, int N = 0, typename Callable>
Fn* fnptr(Callable&& c)
{
    return fnptr_<N>(std::forward<Callable>(c), (Fn*)nullptr);
}

And use it as

void foo(void (*fn)())
{
    fn();   
}

int main()
{
    int i = 42;
    auto fn = fnptr<void()>([i]{std::cout << i;});
    foo(fn);  // compiles!
}

Live

This is essentially declaring an anonymous function at each occurrence of fnptr.

Note that invocations of fnptr overwrite the previously written callable given callables of the same type. We remedy this, to a certain degree, with the int parameter N.

std::function<void()> func1, func2;
auto fn1 = fnptr<void(), 1>(func1);
auto fn2 = fnptr<void(), 2>(func2);  // different function

A simular answer but i made it so you don't have to specify the type of returned pointer (note that the generic version requires C++20):

#include <iostream>


template<typename Function>
struct function_traits;

template <typename Ret, typename... Args>
struct function_traits<Ret(Args...)> {
    typedef Ret(*ptr)(Args...);
};

template <typename Ret, typename... Args>
struct function_traits<Ret(*const)(Args...)> : function_traits<Ret(Args...)> {};

template <typename Cls, typename Ret, typename... Args>
struct function_traits<Ret(Cls::*)(Args...) const> : function_traits<Ret(Args...)> {};

using voidfun = void(*)();

template <typename F>
voidfun lambda_to_void_function(F lambda) {
    static auto lambda_copy = lambda;

    return []() {
        lambda_copy();
    };
}

// requires C++20
template <typename F>
auto lambda_to_pointer(F lambda) -> typename function_traits<decltype(&F::operator())>::ptr {
    static auto lambda_copy = lambda;
    
    return []<typename... Args>(Args... args) {
        return lambda_copy(args...);
    };
}



int main() {
    int num;

    void(*foo)() = lambda_to_void_function([&num]() {
        num = 1234;
    });
    foo();
    std::cout << num << std::endl; // 1234

    int(*bar)(int) = lambda_to_pointer([&](int a) -> int {
        num = a;
        return a;
    });
    std::cout << bar(4321) << std::endl; // 4321
    std::cout << num << std::endl; // 4321
}

Not a direct answer, but a slight variation to use the "functor" template pattern to hide away the specifics of the lambda type and keeps the code nice and simple.

I was not sure how you wanted to use the decide class so I had to extend the class with a function that uses it. See full example here: https://godbolt.org/z/jtByqE

The basic form of your class might look like this:

template <typename Functor>
class Decide
{
public:
    Decide(Functor dec) : _dec{dec} {}
private:
    Functor _dec;
};

Where you pass the type of the function in as part of the class type used like:

auto decide_fc = [](int x){ return x > 3; };
Decide<decltype(decide_fc)> greaterThanThree{decide_fc};

Again, I was not sure why you are capturing x it made more sense (to me) to have a parameter that you pass in to the lambda) so you can use like:

int result = _dec(5); // or whatever value

See the link for a complete example

Here is another variation of the solution. C++14 (can be turned into C++11) Supports return values, non-copyable and mutable lambdas. If mutable lambdas not needed, can be even shorter by removing specialization matching non-const version and embedding impl_impl.

For those who wonder, it works because each lambda is unique (is distinct class) and thus invocation of to_f generates unique for this lambda static and corresponding C-style function which can access it.

template <class L, class R, class... Args> static auto impl_impl(L l) {
  static_assert(!std::is_same<L, std::function<R(Args...)>>::value,
                "Only lambdas are supported, it is unsafe to use "
                "std::function or other non-lambda callables");

    static L lambda_s = std::move(l);
    return +[](Args... args) -> R { return lambda_s(args...); };
}

template <class L>
struct to_f_impl : public to_f_impl<decltype(&L::operator())> {};
template <class ClassType, class R, class... Args>
struct to_f_impl<R (ClassType::*)(Args...) const> {
  template <class L> static auto impl(L l) {
    return impl_impl<L, R, Args...>(std::move(l));
  }
};
template <class ClassType, class R, class... Args>
struct to_f_impl<R (ClassType::*)(Args...)> {
  template <class L> static auto impl(L l) {
    return impl_impl<L, R, Args...>(std::move(l));
  }
};

template <class L> auto to_f(L l) { return to_f_impl<L>::impl(std::move(l)); }

Note, this also tend to work for other callable objects like std::function but it would be better if it didn't work because unlike lambdas, std::function like objects do not generate unique type so inner template and its inner static will be reused for/shared by all functions with same signature, which most probably is not what we want from it. I've specifically disallowed std::function but there exist more which I don't know how to disallow in generic way.

As pointed out by all others capturing lambdas can't be converted to a function pointer, but not the actual reason. Well it's forbidden by the standard, but why they (probably) chose to forbid it.

The reason is that a function pointer(*) is normally implemented as a pointer to (machine) code that when jumped to with parameters put on the stack or wherever the calling convention prescribes will perform the function and place the return value wherever the calling convention prescribe and return to the caller (according how the calling convention says).

There is no place in that implementation to pass the captured values in other ways than just put them in the machine code itself, which would mean having to resort to generating the code (unless the captured values are known at compile time) to be pointed at on the fly (during runtime), which in turn is poorly supported by some platform (and might even be outright forbidden by other). So that wouldn't be a feasible implementation.

And implementing function pointer in another way in order to support converting capturing lambdas to function pointer would break binary compatibility.

So you will inevitably be left with alternatives that aren't function pointers(*).

(*) Meaning C-compatible function pointers at least. Pointer to member function is a little bit more complex, but I don't think it would help anyway.

using DecisionFn = bool(*)();

class Decide
{
public:
    Decide(DecisionFn dec) : _dec{dec} {}
private:
    DecisionFn _dec;
};

int main()
{
    int x = 5;
    static auto lmb = [x](){ return x > 3; };
    Decide greaterThanThree{ [](){ return lmb(); } };
    return 0;
}

https://godbolt.org/z/Mx6qnasvz

A shortcut for using a lambda with as a C function pointer is this:

"auto fun = +[](){}"

Using Curl as exmample (curl debug info)

auto callback = +[](CURL* handle, curl_infotype type, char* data, size_t size, void*){ //add code here :-) };
curl_easy_setopt(curlHande, CURLOPT_VERBOSE, 1L);
curl_easy_setopt(curlHande,CURLOPT_DEBUGFUNCTION,callback);

While the template approach is clever for various reasons, it is important to remember the lifecycle of the lambda and the captured variables. If any form of a lambda pointer is is going to be used and the lambda is not a downward continuation, then only a copying [=] lambda should used. I.e., even then, capturing a pointer to a variable on the stack is UNSAFE if the lifetime of those captured pointers (stack unwind) is shorter than the lifetime of the lambda.

A simpler solution for capturing a lambda as a pointer is:

auto pLamdba = new std::function<...fn-sig...>([=](...fn-sig...){...});

e.g., new std::function<void()>([=]() -> void {...}

Just remember to later delete pLamdba so ensure that you don't leak the lambda memory. Secret to realize here is that lambdas can capture lambdas (ask yourself how that works) and also that in order for std::function to work generically the lambda implementation needs to contain sufficient internal information to provide access to the size of the lambda (and captured) data (which is why the delete should work [running destructors of captured types]).