Re: [Boost-users] fusion::result_of::transform<Seq1, Seq2, F> -- argument dependent result type

13 Sep 2008


      e r wrote:
...
Steven Watanabe wrote:
...
AMDG
e r wrote:
...
// probably not what result_of::transform expects, based on compile
// errors (below)
  template<typename Metafun,typename Arg2>
  struct result{
     typedef typename Metafun::type type;
  };
Fusion uses Boost.ResultOf.
http://www.boost.org/doc/libs/1_36_0/libs/utility/utility.htm#result_of
In Christ,
Steven Watanabe
Thanks. I read the link above before posting as well as the proposal:
//example from proposal.
struct square_ {
  template<typename T> struct result {
    typedef T type;
  }
template<typename T> T operator()(T& x) const
  {
    return x*x;
  }
};
but I'm not qualified enough to deduce the right answer in my case. i
thought id would be like this:
template<typename Arg1>
struct fun{
  template<typename Metafun,typename Arg2>
  struct result{
     typedef typename Metafun::type type;
  };
template<typename Metafun,typename Arg2>
  typename Metafun::type
  operator()(const Metafun& meta,const Arg2& arg2)const{
      typedef typename Metafun::type result_type;
      return result_type(args,arg2);
  }
};
but apparently it's not correct. I guess I can't fully make sense of
F::result<F(T1, T2, ..., TN)>::type. any hint here would be really helpful.
thanks!
I may have made a little bit of progress here...

struct fun{

 template<typename Signature> struct result;

 template<typename A>
 struct result<fun(const A&)>{
    typedef A type;
 };

 template<typename A>
 A operator()(const A& a)const{
    return A();
 };
};

typedef boost::result_of<fun(const my&)>::type result_type;

    BOOST_MPL_ASSERT(
        (
            mpl::is_same<
                result_type,
                my
            >
        )
    );