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From "Eric Lemings" <>
Subject Another potential hole in the tuple specs
Date Mon, 07 Jul 2008 23:14:29 GMT

Here's another potential problem with the tuple spec.  The latest draft
declares std::ignore like so:
	namespace std {
	    const /*unspecified*/ ignore;
The type of std::ignore is implementation-defined but for illustration,
let's say its defined like this:

	namespace std {

	struct _Ignore
	    template <class _Type>
	    _Ignore& operator= (const _Type& value) { return *this; }

	const _Ignore ignore = _Ignore ();

	} // namespace std

(The need for the operator will become evident shortly.)

Here's how the tie() function is specified, quoting from the standard:

	  template<class... Types}
	  tuple<Types&...> tie(Types&... t);

	4 Returns: tuple<Types&>(t...). When an argument in t is ignore,
assigning any value to the corresponding
	  tuple element has no effect.

	5 [ Example: tie functions allow one to create tuples that
unpack tuples into variables. ignore can be used for
	    elements that are not needed:

		int i; std::string s;
		tie(i, ignore, s) = make_tuple(42, 3.14, "C++");
		// i == 42, s == "C++"

	  -end example ]

In the example, the return type of the call to the tie() function is
std::tuple<int&, const std::_Ignore&, std::string&>.  Note that the
second element type in the tuple is a constant reference.  Regardless of
the implementation-defined type of std::ignore, isn't it impossible to
change the value of a constant reference once initialized?  This would
mean the example shown above is ill-formed I believe.


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