C++0x Lambda to function pointer in VS 2010

Question

I am trying to use a lambda to pass in place of a function pointer but VS2010 can't seem to convert it. I have tried using std::function like this and it crashes and I have no idea if I am doing this right!

#include <windows.h>
#include <conio.h>

#include <functional>
#include <iostream>

#include <concrt.h>


void main()
{
    std::function<void(void*)> f = [](void*) -> void
    {
        std::cout << "Hello\n";
    };


    Concurrency::CurrentScheduler::ScheduleTask(f.target<void(void*)>(), 0);

    getch();
}

It seems strange to me that the compiler can't convert such a lambda to a simple function pointer as it captures no variables - also in the case that it did I wonder what can be done.

Is the type of each lambda unique? So I could hack around with a template function using the lambdas' type as a template argument to generate a unique static function that could be called instead and hopefully optimised out?

UPDATED

The below seems to work but is it safe?

#include <windows.h>
#include <conio.h>

#include <iostream>

#include <concrt.h>


template<typename Signature>
struct Bind
{
    static Signature method;

    static void Call(void* parameter)
    {
        method(parameter);
    }
};


template<typename Signature>
Signature Bind<Signature>::method;


template<typename Signature>
void ScheduleTask(Signature method)
{
    Bind<Signature>::method = method;
    Concurrency::CurrentScheduler::ScheduleTask(&Bind<Signature>::Call,0);
}


void main()
{
    ScheduleTask
    (   
        [](void*)
        {
            std::cout << "Hello";
        }
    );


    ScheduleTask
    (   
        [](void*)
        {
            std::cout << " there!\n";
        }
    );


    getch();
}

UPDATED AGAIN

So with the help given I have come up with the shorter:

template<typename Signature>
void (*LambdaBind(Signature))(void*)
{
    struct Detail
    {
        static void Bind(void* parameter)
        {
            Signature method;

            method(parameter);
        }
    };


    return &Detail::Bind;
}

This can be used to wrap a lambda with no closure of void(*)(void*) into the equivalent function pointer. It appears that this will become unnecessary in a later version of VS2010.

So how to get this to work for a lambda with closures?

UPDATED AGAIN!

Works for closures in VS2010 - no idea if it's 'safe' though...

template<typename Signature>
struct Detail2
{
    static std::function<void(void*)> method;


    static void Bind(void* parameter)
    {
        method(parameter);
    }
};


template<typename Signature>
std::function<void(void*)> Detail2<Signature>::method;


template<typename Signature>
void (*LambdaBind2(Signature method))(void*)
{
    Detail2<Signature>::method = method;
    return &Detail2<Signature>::Bind;
}

Ah that helped... and I have a work around now from that which I think is safe in my updated question? — iam
– iam, Commented Jul 28, 2010 at 9:27
Seems safe enough. I'll provide you with a more generic solution. You should remove static Signature method; and instead put Signature method; within the function. It's empty so it won't cost anything anyway. — GManNickG
– GManNickG, Commented Jul 28, 2010 at 9:30

Community · Accepted Answer · 2017-05-23 10:26:39Z

5

This feature of lambda's was added after VS2010 implemented them, so they don't exist in it yet.

Here's a possible generic work-around, very untested:

#include <functional>
#include <iostream>

namespace detail
{
    // helper specializations,
    // define forwarding methods
    template <typename Lambda, typename Func>
    struct lambda_wrapper;

    #define DEFINE_OPERATOR \
            typedef decltype(&call) function_type; \
            operator function_type(void) const \
            { \
                return &call; \
            }

    template <typename Lambda, typename C, typename R>
    struct lambda_wrapper<Lambda, R (C::*)(void) const>
    {
        static R call(void)
        {
            Lambda x;
            return x();
        }

        DEFINE_OPERATOR
    };

    template <typename Lambda, typename C, typename R,
                typename A0>
    struct lambda_wrapper<Lambda, R (C::*)(A0) const>
    {
        static R call(A0&& p0)
        {
            Lambda x;
            return x(std::forward<A0>(p0));
        }

        DEFINE_OPERATOR
    };

    // and so on
    #undef DEFINE_OPERATOR
}

// wraps a lambda and provides 
// a way to call it statically
template <typename Lambda>
struct lambda_wrapper :
        detail::lambda_wrapper<Lambda, decltype(&Lambda::operator())>
{};

template <typename Lambda>
lambda_wrapper<Lambda> wrap_lambda(const Lambda&)
{
    return lambda_wrapper<Lambda>();
}

int main(void)
{
    auto l = [](){ std::cout << "im broked :(" << std::endl; };
    std::function<void(void)> f = wrap_lambda(l);

    f();
}

Let me know if any part is confusing.

edited May 23, 2017 at 10:26

CommunityBot

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answered Jul 28, 2010 at 9:42

GManNickG

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3 Comments

GManNickG Over a year ago

@wb: Heh, decltype just gives you the type of an expression. And && are "rvalue references".

iam Over a year ago

So how can this be made to work for a lambda with a closure? PS: I have updated my question with code for a shorter version for my specific signature case.

GManNickG Over a year ago

@wb: Oops, sorry I missed your comments. Glad you got it solved though. :)

NoNaMe · Accepted Answer · 2012-12-10 04:23:02Z

0

If scheduling lambdas/function objects in Concurrency::CurrentScheduler is what you want, it may be worth your while looking at ConcRT Sample Pack v0.32 here

The task_scheduler struct can schedule lambdas asynchronously, but be advised, passing by reference may cause bad things to happen (since we are talking about asynchronous scheduling without a join/wait, a reference on the stack may no longer be valid at time of task execution!)

edited Dec 10, 2012 at 4:23

NoNaMe

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answered Oct 11, 2010 at 21:36

Vinod

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