Lambda <-> closure equivalence in C++

Question

I was watching a video called Lambda? You Keep Using that Letter by Kevlin Henney, where he states that closures and objects are fundamentaly equivalent:

He then proves his point by this javascript code which implement a stack as a closure:

const newStack = () => {
    const items = []
    return {
        depth: () => items.lengh,
        top: () => items[0],
        push: newTop => { items.unshift(newTop) },
        pop: () => { items.shift() },
    }
}

The advantage of a closure versus a class is that its state is really hidden, whereas a private member is more "inaccessible" than "hidden".

I tried to do something equivalent in C++. However, it seems that it is difficult to express this in C++.

My current version is there, and it has two major drawbacks:

it does compile, but it will not work (the inner shared_ptris released immediately after the closure creation)
is a bit verbose : depth, top, push and pop are repeated 3 times.

auto newStack = []() {
  auto items = std::make_shared<std::stack<int>>();

  auto depth = [&items]() { return items->size();};
  auto top = [&items]() { return items->top(); };
  auto push = [&items](int newTop) {  items->push(newTop); };
  auto pop = [&items]() { items->pop(); };

  struct R {
    decltype(depth) depth;
    decltype(top) top;
    decltype(push) push;
    decltype(pop) pop;
  };
  return R{ depth, top, push, pop};
};

godbolt version here

Is there a working way to do it in C++?

Your program compiles, but doesn't actually work and print out the right results. — cigien
– cigien, Commented May 15, 2020 at 22:01
@cigien: I updated the question in order to mention that the code is actualy buggy — Pascal T.
– Pascal T., Commented May 15, 2020 at 22:17

Jerry Coffin · Accepted Answer · 2020-05-15 22:23:55Z

Yes, of course there's a better way to do it in C++: don't use a lambda.

A lambda expression defines a class. A closure is an instance of that class--an object. We don't need comparisons to other languages to tell us that--it's exactly how lambdas and closures are defined in C++. §[expr.prim.lambda.closure]:

The type of a lambda-expression (which is also the type of the closure object) is a unique, unnamed non-union class type, called the closure type, whose properties are described below.

But (and this is an important point) at least in C++, a lambda expression defines a class with a very limited public interface. Specifically, it provides an overload of operator(), and if it doesn't capture anything, a conversion to pointer to function. If it does capture something, it also defines a constructor to do the capturing. And of course if it captures things, it defines member variables to hold whatever it captures.

But that's all it really defines. It's not that it's doing a better job of hiding whatever else it may contain. It's that it really doesn't contain anything else.

In your case, you're trying to define a type that has four separate member functions that all operate on some state they share. As you've shown, it's sort of possible to do that externalizing the state so you have something nearly equivalent to some C code (or something on that order) that simply has data and some functions that operate on that data. And yes, you can push them together into a structure to do at least some imitation of a class with member functions.

But you're pretty much fighting against the system (so to speak) to do that in C++. Lambdas/closures (as they're defined in C++) are not intended to let you define things that have multiple separate entry points, each carrying out separate actions on shared data. As Samuel Johnson's old line says, "[It] is like a dog walking on its hind legs. It is not done well; but you are surprised to find it done at all."

Collectives™ on Stack Overflow

Lambda <-> closure equivalence in C++

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related