range-based for loop over references

Question

This is question is out of curiosity, not necessity. One way I have found C++11's range based for loop useful is for iterating over discrete objects:

#include <iostream> #include <functional> int main() { int a = 1; int b = 2; int c = 3; // handy: for (const int& n : {a, b, c}) { std::cout << n << '\n'; }

I would like to be able to use the same loop style to modify non-const references too, but I believe it is not allowed by the standard (see Why are arrays of references illegal?):

 // would be handy but not allowed: // for (int& n : {a, b, c}) { // n = 0; // }

I thought of two workarounds but these seem like they could incur some minor additional cost and they just don't look as clean:

 // meh: for (int* n : {&a, &b, &c}) { *n = 0; } // meh: using intRef = std::reference_wrapper<int>; for (int& n : {intRef (a), intRef (b), intRef (c)}) { n = 0; } }

So the question is, is there a cleaner or better way? There may be no answer to this but I'm always impressed with the clever ideas people have on stackoverflow so I thought I would ask.

You could instead just use std::ref. That's about as far as you can get. — Hatted Rooster
– Hatted Rooster, Commented Jun 21, 2018 at 13:04
@SombreroChicken if you put that in an answer I'll accept it, it is an improvement — atb
– atb, Commented Jun 21, 2018 at 13:14

lubgr · Accepted Answer · 2018-06-21 13:57:00Z

Picking up @Sombrero Chicken's idea, here is an approach with less typing:

template <class ...Args> constexpr auto ref(Args&&... args) { using First = std::tuple_element_t<0, std::tuple<Args...>>; using WrappedFirst = std::reference_wrapper<std::remove_reference_t<First>>; constexpr auto n = sizeof...(Args); return std::array<WrappedFirst, n>{std::ref(args)...}; }

which can be used via

for (int& n : ref(a, b, c)) n = 0;

Hatted Rooster · Accepted Answer · 2018-06-21 13:16:57Z

Instead of constructing a reference_wrapper yourself you could use std::ref, that's as far as you can get:

using std::ref; for (int& n : {ref(a), ref(b), ref(c)}) { n = 0; }

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