Re: [boost] determine first arg of callable

Subject: Re: [boost] determine first arg of callable
From: Andrey Semashev (andrey.semashev_at_[hidden])
Date: 2013-05-15 09:35:30

On Wed, May 15, 2013 at 5:13 PM, Oliver Kowalke <oliver.kowalke_at_[hidden]>wrote:

>
> at least what I can do is to overload the ctor:
>
> template< typename Arg >
> X( void(*fn)( Arg &) ) {...}
>
> template< typename Arg >
> X( boost::function< void( Arg &) > fn) {...}
>
> ctor is called then:
>
> X x1( g); // OK
>
> A a;
> boost::function< void( Y< char > &) > f1=boost::bind( & A::operator(), a,
> _1);
> X x2( f1);
>
> boost::function< void( Y< int > &) > f2=boost::bind( g, _1);
> X x3( f2);
>
> For function pointer it's OK - but the code using function<> and bind()
> looks a little bit clumsy:
>
> X x2(boost::bind( & A::operator(), a, _1)); // will not compile

The last one is equivalent to just X x2(a);

You could try something like this:

class X
{
public:
  template< typename R, typename A0 >
  X(R (*p)(A0)) {}

  template< typename R, typename A0, typename A1 >
  X(R (*p)(A0, A1)) {}

  ...

  template< typename R, typename C, typename A0 >
  X(R (C::*p)(A0)) {}

  template< typename R, typename C, typename A0, typename A1 >
  X(R (C::*p)(A0, A1)) {}

  ...

  template< typename T >
  X(T f) : X(&T::operator()) {}
};

This will not work with templated or overloaded operator() (and for this
reason will not work with bind).

But I agree that you should avoid this design.