Passing a non-copyable closure object to std::function parameter

Question

In C++14, a lambda expression can capture variables by moving from them using capture initializers. However, this makes the resulting closure object non-copyable. If I have an existing function that takes a std::function argument (that I cannot change), I cannot pass the closure object, because std::function's constructor requires the given functor to be CopyConstructible.

#include <iostream>
#include <memory>

void doit(std::function<void()> f) {
    f();
}

int main()
{
    std::unique_ptr<int> p(new int(5));
    doit([p = std::move(p)] () { std::cout << *p << std::endl; });
}

This gives the following errors:

/usr/bin/../lib/gcc/x86_64-linux-gnu/4.8/../../../../include/c++/4.8/functional:1911:10: error: 
      call to implicitly-deleted copy constructor of '<lambda at test.cpp:10:7>'
            new _Functor(*__source._M_access<_Functor*>());
                ^        ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
/usr/bin/../lib/gcc/x86_64-linux-gnu/4.8/../../../../include/c++/4.8/functional:1946:8: note: in
      instantiation of member function 'std::_Function_base::_Base_manager<<lambda at test.cpp:10:7>
      >::_M_clone' requested here
              _M_clone(__dest, __source, _Local_storage());
              ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/4.8/../../../../include/c++/4.8/functional:2457:33: note: in
      instantiation of member function 'std::_Function_base::_Base_manager<<lambda at test.cpp:10:7>
      >::_M_manager' requested here
            _M_manager = &_My_handler::_M_manager;
                                       ^
test.cpp:10:7: note: in instantiation of function template specialization 'std::function<void
      ()>::function<<lambda at test.cpp:10:7>, void>' requested here
        doit([p = std::move(p)] () { std::cout << *p << std::endl; });
             ^
test.cpp:10:8: note: copy constructor of '' is implicitly deleted because field '' has a deleted
      copy constructor
        doit([p = std::move(p)] () { std::cout << *p << std::endl; });
              ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/4.8/../../../../include/c++/4.8/bits/unique_ptr.h:273:7: note: 
      'unique_ptr' has been explicitly marked deleted here
      unique_ptr(const unique_ptr&) = delete;
      ^

Is there a reasonable workaround?

Testing with Ubuntu clang version 3.5-1~exp1 (trunk)

Answer 1

There is this approach:

template< typename signature >
struct make_copyable_function_helper;
template< typename R, typename... Args >
struct make_copyable_function_helper<R(Args...)> {
  template<typename input>
  std::function<R(Args...)> operator()( input&& i ) const {
    auto ptr = std::make_shared< typename std::decay<input>::type >( std::forward<input>(i) );
    return [ptr]( Args... args )->R {
      return (*ptr)(std::forward<Args>(args)...);
    };
  }
};

template< typename signature, typename input >
std::function<signature> make_copyable_function( input && i ) {
  return make_copyable_function_helper<signature>()( std::forward<input>(i) );
}

where we make a shared pointer to our data, then make a copyable lambda that captures that shared pointer, then we wrap that copyable lambda into a std::function of the requested signature.

In your case above, you'd just:

doit( make_copyable_function<void()>( [p = std::move(p)] () { std::cout << *p << std::endl; } ) );

A slightly more advanced version defers the type erasure and adds a layer of perfect forwarding to reduce overhead:

template<typename input>
struct copyable_function {
  typedef typename std::decay<input>::type stored_input;
  template<typename... Args>
  auto operator()( Args&&... args )->
    decltype( std::declval<input&>()(std::forward<Args>(args)...) )
  {
    return (*ptr)(std::forward<Args>(args));
  }
  copyable_function( input&& i ):ptr( std::make_shared<stored_input>( std::forward<input>(i) ) ) {}
  copyable_function( copyable_function const& ) = default;
private:
  std::shared_ptr<stored_input> ptr;
};
template<typename input>
copyable_function<input> make_copyable_function( input&& i ) {
  return {std::forward<input>(i)}; 
}

which does not require you to pass the signature in, and can be slightly more efficient in a few cases, but uses more obscure techniques.

In C++14 with this can be made even more brief:

template< class F >
auto make_copyable_function( F&& f ) {
  using dF=std::decay_t<F>;
  auto spf = std::make_shared<dF>( std::forward<F>(f) );
  return [spf](auto&&... args)->decltype(auto) {
    return (*spf)( decltype(args)(args)... );
  };
}

doing away with the need for the helper type entirely.

Answer 2

If lifetime of the closure object isn't an issue, you could pass it in a reference wrapper:

int main()
{
    std::unique_ptr<int> p(new int(5));
    auto f = [p = std::move(p)]{
        std::cout << *p << std::endl;
    };
    doit(std::cref(f));
}

This obviously doesn't apply to every scenario, but it's fine for your example program.

EDIT: Taking a glance at N3797 (C++14 working draft) § 20.9.11.2.1 [func.wrap.func.con] p7, the CopyConstructible requirement is still there. I wonder if there's a technical reason that can't be loosened to MoveConstructible, or if the committee just didn't get around to it?

EDIT: Answering my own question: std::function is CopyConstructible, so the wrapped functor needs to be CopyConstructible as well.

Answer 3

If you know you aren't actually going to copy your function object then you can just wrap it in a type that makes the compiler think it's copyable:

struct ThrowOnCopy {
  ThrowOnCopy() = default;
  ThrowOnCopy(const ThrowOnCopy&) { throw std::logic_error("Oops!"); }
  ThrowOnCopy(ThrowOnCopy&&) = default;
  ThrowOnCopy& operator=(ThrowOnCopy&&) = default;
};

template<typename T>
  struct FakeCopyable : ThrowOnCopy
  {
    FakeCopyable(T&& t) : target(std::forward<T>(t)) { }

    FakeCopyable(FakeCopyable&&) = default;

    FakeCopyable(const FakeCopyable& other)
    : ThrowOnCopy(other),                             // this will throw
      target(std::move(const_cast<T&>(other.target))) // never reached
    { }

    template<typename... Args>
      auto operator()(Args&&... a)
      { return target(std::forward<Args>(a)...); }

    T target;
  };


template<typename T>
  FakeCopyable<T>
  fake_copyable(T&& t)
  { return { std::forward<T>(t) }; }


// ...

doit( fake_copyable([p = std::move(p)] () { std::cout << *p << std::endl; }) );

The function template fake_copyable creates a wrapper which is CopyConstructible according to the compiler (and <type_traits>) but cannot be copied at run-time.

If you store a FakeCopyable<X> in a std::function and then end up copying the std::function you will get a std::logic_error thrown, but if you only move the std::function everything will work OK.

The target(std::move(const_cast<T&>(other.target))) looks worrying, but that initializer will never run, because the base class initializer will throw first. So the worrying const_cast never really happens, it just keeps the compiler happy.