Workaround to allow tr1::function to swallow return values

Question

As a follow-up to Can tr1::function swallow return values?, how can one work around the limitation that tr1::function cannot swallow return values?

This should work in the specific case of swallowing the return value of a callable object taking no arguments:

template<typename FuncT>
struct swallow_return_t
{
  explicit swallow_return_t(FuncT i_Func):m_Func(i_Func){}
  void operator()(){ m_Func(); }
  FuncT m_Func;
};

template<typename FuncT>
swallow_return_t<FuncT>
swallow_return(FuncT f)
{
  return swallow_return_t<FuncT>(f);
}

Then use like:

int Foo();
std::tr1::function<void()> Bar = swallow_return(Foo);

I assume variadic templates and perfect forwarding would permit generalization of this technique to arbitrary parameter lists. Is there a better way?

Answer 1

The following works for me in GCC 4.6.1:

#include <functional>

int foo() { return 5; }
int goo(double, char) { return 5; }

int main()
{
  std::function<void()> f = foo;
  std::function<void(float, int)> g = goo;
  (void)f();
  (void)g(1.0f, 'a');
}

---

Here's a wrapper using lambdas, but it's not automagic yet

template <typename T, typename ...Args>
struct strip_return
{
  static inline std::function<void(Args...)> make_function(std::function<T(Args...)> f)
  {
    return [&f](Args... args) -> void { f(args...); };
  }
};

int main()
{
  auto q = strip_return<int>::make_function(std::bind(foo));
  q();
}

---

Forget the middle part. OK, since std::function is type-erasing, it's hard to get at the underlying types. However, if you go for the function reference directly, you can avoid these problems entirely:

template <typename T, typename ...Args>
static inline std::function<void(Args...)> make_direct(T (&f)(Args...))
{
  return [&f](Args... args) -> void { f(args...); };
}

int main()
{
  auto p = make_direct(foo);
  q();
}