std::function variable arguments in one vector/map

Question

how could one do this in c++ today without using two separate holders?

typedef std::function<void(int a, int b)> f1; 
typedef std::function<void(int a)> f2; 

std::vector<f1> m;

void add(f1 f)
{
    m.push_back(f);
}

void add(f2 f)
{
    // add one more (unused) parameter to f2 so we can add f2 to f1 vector holder?
}

can we somehow overload f1 function to include different set of parameters? could this be solved by variadic templates nowdays or something similar?

Answer 1

Create a new lambda matching the new signature and add that instead:

void add(f2 f)
{
    m.push_back( [g = std::move(f)](int a, int /* unused */){ g(a); } );
}

Answer 2

This wraps a std::function in a lambda that ignores any extra args:

template<class R, class...Args>
auto ignore_extra_args( std::function<R(Args...)> f ) {
  return [f = std::move(f)](Args...args, auto&&...)->R{
    return f(std::forward<Args>(args)...);
  };
}

it gets trickier if we don't want that needless layer of type erasure.

You need to find the longest prefix of Args... which you can invoke a given object f with, then invoke f with it. This involves some tricky metaprogramming.

It is far easier if you ask the caller to pass in a signature:

template<class Sig>
struct ignore_extra_args_helper;
template<class R, class...Args>
struct ignore_extra_args_helper<R(Args...)> {
  template<class F>
  auto operator()( F&& f ) {
    return [f = std::forward<F>(f)](Args...args, auto&&...)->R{
      return f(std::forward<Args>(args)...);
    };
  }
};
template<class Sig, class F>
auto ignore_extra_args( F&& f ) {
  return ignore_extra_args_helper<Sig>{}(std::forward<F>(f));
}

which saves on possible overhead.

template<class F, decltype(std::declval<F const&>()(1,1))* =nullptr>
void add(F&& f) {
  m.push_back(std::forward<F>(f));
}

template<class F, class...Unused>
void add(F&& f, Unused&&...) {
  add( ignore_extra_args<void(int)>(std::forward<F>(f)) );
}

live example