Remove reference from std::tuple members

Question

I'm implementing save/restore functionality for some variables with the help of stl tuples as follows:

double a = 1, b = 2; int c = 3; auto tupleRef = std::make_tuple(std::ref(a), std::ref(b), std::ref(c)); // here I'm saving current state of a, b, c std::tuple<double, double, int> saved = tupleRef; //here goes block of code, where a, b, and c get spoiled ...................... // //now I'm restoring initial state of a, b, c tupleRef = savedTuple;

This code works well. But instead of explicit specifying tuple members types in

std::tuple<double, double, int> saved = tupleRef;

I'd like to rather remove references from all tupleRef members, like in the following

auto saved = remove_ref_from_tuple_members(tupleRef);

I believe that it is possible to write "remove_ref_from_tuple_members" template to that end.

Thanks for answers.

R. Martinho Fernandes · Accepted Answer · 2012-10-05 09:03:41Z

8

A simple type alias can be used to apply std::remove_reference to all types in a tuple.

template <typename... T> using tuple_with_removed_refs = std::tuple<typename std::remove_reference<T>::type...>;

Armed with this you can now write the function template:

template <typename... T> tuple_with_removed_refs remove_ref_from_tuple_members(std::tuple<T...> const& t) { return tuple_with_removed_refs { t }; }

answered Oct 5, 2012 at 9:03

R. Martinho Fernandes

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4 Comments

Steve Jessop Over a year ago

Wait, you mean template parameter packs have a built-in "map" operation (in the functional programming sense, not the std::map sense)? Praise be, that I should live to see such times!

R. Martinho Fernandes Over a year ago

@SteveJessop Yes, I think you can say that. You can expand a pack into any pattern.

Alex B. Over a year ago

Thanks for the answer. I believe it works, at least looks logical. Unfortunately variadic templates don’t compile on my computer in Visual Studil 2010 nor in VS2012, they are just not supported by VS. It looks like they hard-coded tuple to have max 15 members.

M.M Over a year ago

To fix it I had to change both instances of tuple_with_removed_refs in the second function to tuple_with_removed_refs<T...>

Alex B. · Accepted Answer · 2012-10-08 08:57:56Z

Thanks to R. Martinho Fernandes, whose code I was able to modify to compile in Visual Studio where tuple is hard-coded as a template with 10 types with unused types being empty structs.

#define _RR_(x) typename std::remove_reference<x>::type #define _no_ref_tuple_ std::tuple<_RR_(T0), _RR_(T1), _RR_(T2), _RR_(T3), _RR_(T4), _RR_(T5), _RR_(T6), _RR_(T7), _RR_(T8), _RR_(T9)> template <typename T0, typename T1, typename T2, typename T3, typename T4, typename T5, typename T6, typename T7, typename T8, typename T9> _no_ref_tuple_ map_remove_ref(std::tuple<T0, T1, T2, T3, T4, T5, T6, T7, T8, T9> const& t) { return _no_ref_tuple_(t); }

I also think that binding of the refs into tuple like in

auto tupleRef = std::make_tuple(std::ref(x_0), ..., std::ref(x_n));

can be made less verbose:

auto tupleRef = std::forward_as_tuple(x_0, ..., x_n);

but this again will not work in VS, as there is no std::forward_as_tuple.

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