> -----Original Message----- > From: Martin Sebor [mailto:[EMAIL PROTECTED] On Behalf Of Martin Sebor > Sent: Monday, July 07, 2008 5:44 PM > To: dev@stdcxx.apache.org > Subject: Re: Another potential hole in the tuple specs > > Eric Lemings wrote: > > > > > > 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. > > You mean because tie(i, ignore, s) = make_tuple(42, 3.14, "C++") > must assign 3.14 to std::ignore? Would declaring the assignment > operator cost be a way to make it work?
A const assignment operator? Sounds unorthodox but I'll try it out. My current workaround is to declare std::ignore mutable (i.e. non-const). A const assignment operator (if it works) would be preferable; no visible workaround required. Brad.
