How to convert a lambda to an std::function using

问题:

Basically, what I want to be able to do is take a lambda with any number of any type of parameters and convert it to an std::function. I\'ve tried the following and neither method works.

std::function([](){});//Complains that std::function is missing template parameters
template <typename T> void foo(function<T> f){}
foo([](){});//Complains that it cannot find a matching candidate

The following code does work however, but it is not what I want because it requires explicitly stating the template parameters which does not work for generic code.

std::function<void()>([](){});

I\'ve been mucking around with functions and templates all evening and I just can\'t figure this out, so any help would be much appreciated.

As mentioned in a comment, the reason I\'m trying to do this is because I\'m trying to implement currying in C++ using variadic templates. Unfortunately, this fails horribly when using lambdas. For example, I can pass a standard function using a function pointer.

template <typename R, typename...A>
void foo(R (*f)(A...)) {}
void bar() {}
int main() {
    foo(bar);
}

However, I can\'t figure out how to pass a lambda to such a variadic function. Why I\'m interested in converting a generic lambda into an std::function is because I can do the following, but it ends up requiring that I explicitly state the template parameters to std::function which is what I am trying to avoid.

template <typename R, typename...A>
void foo(std::function<R(A...)>) {}
int main() {
    foo(std::function<void()>([](){}));
}

回答1:

You can\'t pass a lambda function object as an argument of type std::function<T> without explicitly specifying the template argument T. Template type deduction tries to match the type of your lambda function to the std::function<T> which it just can\'t do in this case - these types are not the same. Template type deduction doesn\'t consider conversions between types.

It is possible if you can give it some other way to deduce the type. You can do this by wrapping the function argument in an identity type so that it doesn\'t fail on trying to match the lambda to std::function (because dependent types are just ignored by type deduction) and giving some other arguments.

template <typename T>
struct identity
{
  typedef T type;
};

template <typename... T>
void func(typename identity<std::function<void(T...)>>::type f, T... values) {
  f(values...);
}

int main() {
  func([](int x, int y, int z) { std::cout << (x*y*z) << std::endl; }, 3, 6, 8);
  return 0;
}

This is obviously not useful in your situation though because you don\'t want to pass the values until later.

Since you don\'t want to specify the template parameters, nor do you want to pass other arguments from which the template parameters can be deduced, the compiler won\'t be able to deduce the type of your std::function argument.

回答2:

You can use a dedicated/retrospective cast. Once you have a tool like this

#include <functional>

using namespace std;

template<typename T>
struct memfun_type
{
    using type = void;
};

template<typename Ret, typename Class, typename... Args>
struct memfun_type<Ret(Class::*)(Args...) const>
{
    using type = std::function<Ret(Args...)>;
};

template<typename F>
typename memfun_type<decltype(&F::operator())>::type
FFL(F const &func)
{ // Function from lambda !
    return func;
}

you can say FFL() to all lambda types to have them converted to what would be the correct version of std::function

template <typename... Args> void Callback(std::function<void(Args...)> f){
    // store f and call later
}

int main()
{
    Callback(FFL([](int a, float b){
        // do something
    }));

    return 0;
}

Display

回答3:

As shown at Inferring the call signature of a lambda or arbitrary callable for "make_function", you can infer the calling signature of a lambda (or any other functor with a single calling signature) from its (single) operator():

template<typename T> struct remove_class { };
template<typename C, typename R, typename... A>
struct remove_class<R(C::*)(A...)> { using type = R(A...); };
template<typename C, typename R, typename... A>
struct remove_class<R(C::*)(A...) const> { using type = R(A...); };
template<typename C, typename R, typename... A>
struct remove_class<R(C::*)(A...) volatile> { using type = R(A...); };
template<typename C, typename R, typename... A>
struct remove_class<R(C::*)(A...) const volatile> { using type = R(A...); };

template<typename T>
struct get_signature_impl { using type = typename remove_class<
    decltype(&std::remove_reference<T>::type::operator())>::type; };
template<typename R, typename... A>
struct get_signature_impl<R(A...)> { using type = R(A...); };
template<typename R, typename... A>
struct get_signature_impl<R(&)(A...)> { using type = R(A...); };
template<typename R, typename... A>
struct get_signature_impl<R(*)(A...)> { using type = R(A...); };
template<typename T> using get_signature = typename get_signature_impl<T>::type;

This is a rather inflexible approach, though; as R. Martinho Fernandes says, it won\'t work for functors with multiple operator()s, nor for functors with templated operator() or for (C++14) polymorphic lambdas. This is why bind defers inference of its result type until the eventual call attempt.

回答4:

It is possible to get the needed std::function type for lambda using derivation, decltype, variadic templates and a few type traits:

namespace ambient {

    template <typename Function>
    struct function_traits : public function_traits<decltype(&Function::operator())> {};

    template <typename ClassType, typename ReturnType, typename... Args>
    struct function_traits<ReturnType(ClassType::*)(Args...) const> {
        typedef ReturnType (*pointer)(Args...);
        typedef const std::function<ReturnType(Args...)> function;
    };

    template <typename Function>
    typename function_traits<Function>::function to_function (Function& lambda) {
        return static_cast<typename function_traits<Function>::function>(lambda);
    }

    template <class L>
    struct overload_lambda : L {
        overload_lambda(L l) : L(l) {}
        template <typename... T>
        void operator()(T&& ... values){
            // here you can access the target std::function with
            to_function(*(L*)this)(std::forward<T>(values)...);
        }
    };

    template <class L>
    overload_lambda<L> lambda(L l){
        return overload_lambda<L>(l);
    }

}

I use it in my code like this:

ambient::lambda([&](const vector<int>& val){ // some code here // })(a);

PS: in my real case I then save this std::function object and its arguments inside a generic kernel objects that I can execute later on demand via virtual functions.

     

回答5:

Isn\'t currying already implemented with std::bind?

auto sum = [](int a, int b){ return a+b; };
auto inc = std::bind( sum, _1, 1 );
assert( inc(1)==2 );

回答6:

This could be interesting for you: https://gist.github.com/Manu343726/94769034179e2c846acc

That is an experiment I have written a month ago. The goal was to create a functor-like C++ template which emulates Haskell\'s partial calls closures, i.e. the automatic creation of a closure of m-n argumments when you call with n argumments a function with m parameters.

This is one example of what this experiment is cappable to do:

int f( int a, int b, int c, int d)
{
    return a+b+c+d;
}

int main()
{
    auto foo = haskell::make_function( f );

    auto a = foo , 1 , 2 , 3; //a is a closure function object with one parameter

    std::cout << a , 4 << std::endl; //Prints 10
}

haskell::make_function uses some type traits to take care of the different types of function entities, lambdas included:

auto f = haskell::make_function( []( int x, int y , int z ){ return x*y*z; } );

auto a = f(1,2); //a is functor with one parameter (Using the alternative C++-like syntax)
auto b = a(3); // b is 6

As you can see, I use comma operator to mmimic Haskell syntax, but you could change it to the call operator to achieve your goal syntax.

Your are completely free to do anything you want with the code (Check the license).