Re: [Boost-users] [boost::function] problem with void type as argument type

From: Douglas Gregor (doug.gregor_at_[hidden])
Date: 2007-03-26 15:46:33

On Mon, 2007-03-26 at 21:39 +0200, Seweryn Habdank-Wojewódzki wrote:
> Hi!
> 
> Douglas Gregor wrote:
> 
> >> template < typename T, typename U >
> >> void test ( boost::function<T(U)> & f );
> > [snip]
> >>     boost::function<int (void)> h = &f3;
> >>     boost::function<void (void)> i = &f4;
> > 
> > The problem is that "int(void)" is just an alternative spelling for
> > "int()", i.e., a function type for a function that takes no parameters
> > but returns an "int". A function type int() cannot match a template
> > parameter T(U), because there is nothing to match the "U".
> 
> Thanks for reply. However still I do not know how to proceed in that case,
> because functions can not be partially specialized. Is there any
> opportunity to prepare more less such a function which will automatically
> detects arguments of boost::function template?

You can write multiple overloads of your "test" function, each of which
accepts boost::function instances with a different number of function
arguments:

  template<typename R>
  void test (boost::function<T(void)>& f);

  template<typename R, typename T1>
  void test (boost::function<T(T1)>& f);

  template<typename R, typename T1, typename T2>
  void test (boost::function<T(T1, T2)>& f);

  template<typename R, typename T1, typename T2, typename T3>
  void test (boost::function<T(T1, T2, T3)>& f);

Tedious, unfortunately, but it works.

  Cheers,
  Doug