Re: Conversion operator that must convert return...

From:
Gil <vnicula@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Tue, 30 Aug 2011 15:24:36 -0700 (PDT)
Message-ID:
<51cdaa97-b7a9-4f9e-bcda-eadca0b14755@n19g2000prh.googlegroups.com>
On Aug 30, 1:13 pm, Noah Roberts <roberts.n...@gmail.com> wrote:

I am trying to do something like so:

std::tie(x,y) = object;

My initial attempt was something like so:

struct attempt0
{
  template < typename ... T >
  operator std::tuple<T...> ()
  {
    return std::tuple<T...>();
  }

};

The problem here is that tie returns a tuple of reference types, I
can't just construct and return an empty one. It is assignable from
tuples that have matching non-reference types though so my next naive
attempt was:

struct attempt1
{
  template < typename ... T >
  operator std::tuple<typename std::remove_reference<T>::type...>()
  {
    return std::tuple<typename std::remove_reference<T>::type...>();
  }

};

This version is of course not recognized though. Third attempt was:

struct attempt2
{
  template < typename ... T >
  operator std::tuple<T...>()
  {
    return std::tuple<typename std::remove_reference<T>::type...>();
  }

};

I didn't expect this to work and it of course didn't.

Sitting here trying to come up with legal C++ to do what I'm trying
and I can't think of any. Any ideas?


you have to *add* reference to match the tie decl type... see below.

/**
 * @file: tie_test.cpp
 * @author: gil
 * Distributed under the Boost Software License, Version 1.0.
 */

#include <tuple>
#include <iostream>
#include <functional>

template< typename X, typename Y >
struct object {
  object( X const & x, Y const & y ) : x( x ), y( y ) { }
  operator std::tuple< X&, Y& >() {
    return std::make_tuple( std::ref( x ), std::ref( y ) );
  }
  X x; Y y;
};

int main( ) {

  object< int, int > c0( 0, 1 );
  int i, d;
  std::tie( i, d ) = c0; /* problem solved */
  std::cout << i << " " << d << std::endl;

}

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