Instantiating C++ lambda by its type

Question

I want a way to make functor from function. Now I trying to wrap function call by lambda function and instantiate it later. But compiler says than lambda constructor is dele

Answer 1

Lambdas do not have default constructors. Ever. The only constructors they sometimes give access to is (depending on what they capture) the copy and/or move constructors.

If you create a functor with no public default constructor, you'll get the same error.

In C++17 you can solve this with constexpr lambda and operator+ to decay-to-function pointer. A type that carries a function pointer and invokes it is easy with auto template parameters.

In C++11 you gotta get a bit hacky.

template
struct stateless_lambda_t {
  static std::aligned_storage_t< sizeof(F), alignof(F) >& data() {
    static std::aligned_storage_t< sizeof(F), alignof(F) > retval;
    return retval;
  };
  template, stateless_lambda_t >{}, int> =0
  >
  stateless_lambda_t( Fin&& f ) {
    new ((void*)&data()) F( std::forward(f) );
  }
  stateless_lambda_t(stateless_lambda_t const&)=default;
  template
  decltype(auto) operator()(Args&&...args)const {
    return (*static_cast( (void*)&data() ))(std::forward(args)...);
  }
  stateless_lambda_t() = default;
};
template
stateless_lambda_t> make_stateless( F&& fin ) {
  return {std::forward(fin)};
}

Now we can:

auto t = make_stateless([]{ func(); });

and your code works.

A static_assert or SFINAE that the F is actually an empty type might be a good idea. You know, for quality.

The use of C++14 features can be replaced with manual decltype and spewing typename and ::type keywords. This answer was originally written for a C++14 question that was closed as a duplicate of this one.

live example.