Re: Conversion operator that must convert return...

From:

Victor Bazarov <v.bazarov@comcast.invalid>

Newsgroups:

comp.lang.c++

Date:

Tue, 30 Aug 2011 14:19:04 -0400

Message-ID:

<j3j9ip$ugo$1@dont-email.me>

On 8/30/2011 1:13 PM, Noah Roberts 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?

Not that I know much, but it seems that you're trying to guess what your
object is going to be assigned to. That's impossible. Perhaps you
should consider some other syntax altogether. What problem are you
trying to solve?

BTW, a usual solution when you need to return an "empty" reference is to
have a static object [in your class] a reference to which you can return
(and check against) in case you need "an empty one".

V
--
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