CODEWITHSUNDEEP
c++c++11macrosundefined-behaviorforwarding-reference
Claudiu
Claudiu

Reputation: 229451

How can I create a macro which uses a value multiple times, without copying it?

I'd like to create a macro which unpacks a pair into two local variables. I'd like to not create a copy of the pair if it's just a variable, which this would accomplish:

#define UNPACK_PAIR(V1, V2, PAIR) \
    auto& V1 = PAIR.first; \
    auto& V2 = PAIR.second;

UNPACK_PAIR(one, two, x);

However, I'd also like it to not evaluate the expression it's given multiple times, e.g. this should only call expensive_computation() once:

UNPACK_PAIR(one, two, expensive_computation());

If I do:

#define UNPACK_PAIR_A(V1, V2, PAIR) \
    auto tmp = PAIR; \
    auto& V1 = tmp.first; \
    auto& V2 = tmp.second;

then it works for the expensive_computation() case, but it makes a copy in the x case. If I do:

#define UNPACK_PAIR_R(V1, V2, PAIR) \
    auto& tmp = PAIR; \
    auto& V1 = tmp.first; \
    auto& V2 = tmp.second;

Then it works in the x case without making a copy but fails in the expensive_computation() case. If I do:

#define UNPACK_PAIR_CR(V1, V2, PAIR) \
    const auto& tmp = PAIR; \
    auto& V1 = tmp.first; \
    auto& V2 = tmp.second;

#define UNPACK_PAIR_RR(V1, V2, PAIR) \
    auto&& tmp = PAIR; \
    auto& V1 = tmp.first; \
    auto& V2 = tmp.second;

These both compile and run, but I suspect they invoke undefined behavior - am I correct about that? Also, would either of these make any sense?

#define UNPACK_PAIR_RR(V1, V2, PAIR) \
    auto&& tmp = std::move(PAIR); \
    auto& V1 = tmp.first; \
    auto& V2 = tmp.second;

#define UNPACK_PAIR_RR(V1, V2, PAIR) \
    auto&& tmp = std::forward<decltype(PAIR)>(PAIR); \
    auto& V1 = tmp.first; \
    auto& V2 = tmp.second;

Is there any way to create a macro that works for both of these use cases - not copying x yet also not invoking undefined behavior when given the result of an expression or function call?

Upvotes: 6

Views: 450

Answers (3)

celticminstrel
celticminstrel

Reputation: 1679

What you want is std::tie.

decltype(p.first) x;
decltype(p.second) y;
std::tie(x,y) = p;

If you want, you could even use that to define your macro. Note that this will only work for 2-tuples - if you want 3-tuples or more, you'll need to do it a bit differently. For example, if you have a 3-tuple t:

decltype(std::get<0>(t)) x;
decltype(std::get<1>(t)) y;
decltype(std::get<2>(t)) z;
std::tie(x,y,z) = t;

Upvotes: 2

Angew is no longer proud of SO
Angew is no longer proud of SO

Reputation: 171167

auto&& creates a forwarding reference, i.e. it accepts anything. It does not (always) create an rvalue reference. So just do this:

#define UNPACK_PAIR(V1, V2, PAIR) \
    auto&& tmp = PAIR; \
    auto& V1 = tmp.first; \
    auto& V2 = tmp.second;

However, I would strongly suggest against this (unless the scope of the use of UNPACK_PAIR is very limited and the operation is really ubiquitous in that scope). It looks like obscurity for no real benefit to me. Imagine returning to the project after 6 months, with just two hours to find a critical bug. Will you be thanking yourself for using a nonstandard macro-based syntax instead of something readable?

Upvotes: 4

Lightness Races in Orbit
Lightness Races in Orbit

Reputation: 385254

You don't need a macro for this.

auto p = std::make_pair(2, 3);
int x, y;
std::tie(x, y) = p;

If you want references to existing members of a pair:

auto p = std::make_pair(2, 3);
auto& x = p.first;
auto& y = p.second;

That's it.

Now you can move on to something more challenging/interesting/important.

Upvotes: 6

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