C++ conditional reference to existing or new variable

Question

I have a function that forwards some large variables via const reference like this:

void processMore(const std::vector<int> &a, const std::vector<int> &b);

void process(const std::vector<int> &a,
             const std::vector<int> &b) {
  processMore(a, b);
}

But I have added an optimisation that modifies them:

std::pair<std::vector<int>, std::vector<int>>
optimise(const std::vector<int> &a, const std::vector<int> &b);

I can do it unconditionally like this:

void process(const std::vector<int> &a,
             const std::vector<int> &b) {
  auto opt = optimise(a, b);
  processMore(opt.first, opt.second);
}

But I want to make it optional. My solution is this:

void process(const std::vector<int> &a,
             const std::vector<int> &b,
             bool doOptimisation) {
  auto opt =
      [&]() -> std::pair<const std::vector<int>&,
                         const std::vector<int>&> {
    if (doOptimisation) {
      auto o = optimise(a, b);
      return std::make_pair(o.first, o.second);
    }
    return std::make_pair(a, b);
  }();
  processMore(opt.first, opt.second);
}

This seems a bit convoluted. Is there a better way? I want to avoid unnecessary copying.

Red alert! Your lambda returns a pair of dangling references that refer to the dead o! — Quentin
– Quentin, Commented Apr 16, 2019 at 9:45
with "unnecessary" do you mean in case you dont do the optimization or also in case you do it? — 463035818_is_not_an_ai
– 463035818_is_not_an_ai, Commented Apr 16, 2019 at 9:45
With your current lambda, why not just simply return optimise(a, b);? — Some programmer dude
– Some programmer dude, Commented Apr 16, 2019 at 9:45
@Quentin: Ah yeah I did think about that but was thinking that const references would extend the lifetime. I guess that doesn't extend outside the lambda. — Timmmm
– Timmmm, Commented Apr 16, 2019 at 9:48
@Someprogrammerdude: Because optimise() returns a pair of values, and I need a pair of references. — Timmmm
– Timmmm, Commented Apr 16, 2019 at 9:49

Some programmer dude · Accepted Answer · 2019-04-16 09:44:14Z

3

The "naive" solution would be something like

if (doOptimisation)
{
    auto o = optimise(a, b);
    processMore(o.first, o.second);
}
else
{
    processMore(a, b);
}

No extra copying being made.

answered Apr 16, 2019 at 9:44

Some programmer dude

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4 Comments

Timmmm Over a year ago

Ah yeah I guess I should say that processMore() is actually quite a lot of code, so I don't really want to write it all twice.

463035818_is_not_an_ai Over a year ago

@Timmm it is a function that you simply call twice. If it isnt in you real code then the solution is to use a function ;)

Timmmm Over a year ago

Ah I see what you mean. I mean, when I said processMore() is a lot of code I didn't mean the code in processMore() is a lot of code - I meant that the code in process() is a lot longer than just calling one function. But I guess I can put all that code in another function.

John Zwinck Over a year ago

@Timmmm: Exactly, but that's effectively what your lambda is. This question is sort of moot because it was framed in a way that doesn't match your real code.

Timmmm · Accepted Answer · 2019-04-16 12:38:47Z

0

Another option is this:

void process(const std::vector<int> &a,
             const std::vector<int> &b,
             bool doOptimisation) {
  std::optional<std::pair<std::vector<int>, std::vector<int>>> opt;
  if (doOptimisation) {
    opt = optimise(a, b);
  }
  const auto &aOpt = opt ? opt->first : a;
  const auto &bOpt = opt ? opt->second : b;

  processMore(aOpt, bOpt);
}

This is simpler than repeating processMore() or moving it into another wrapper function in my actual code (it's hard to convey complicated code in a minimal example).

answered Apr 16, 2019 at 12:38

Timmmm

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