Should I pass a lambda by const reference.

Question

Typically I use the following pattern when accepting a lambda as an argument to a function (A template class passed-by-value):

template <class Function>
void higherOrderFunction(Function f) {
     f();
}

Does this copy (the closure of) the argument? If so, is there anything wrong with accepting the lambda by const reference instead?

template <class Function>
void higherOrderFunction(const Function& f) {
  f();
}

A simple test seems to indicate that this works fine, but I want to know if there are any special considerations that I should be aware of.

Answer 1

If you pass by value you will copy the closure object (assuming you don't define the lambda inline, in which case it will be moved). This might be undesirable if the state is expensive to copy, and will fail to compile if the state is not copyable.

template <class Function>
void higherOrderFunction(Function f);

std::unique_ptr<int> p;
auto l = [p = std::move(p)] {}; // C++14 lambda with init capture
higherOrderFunction(l);         // doesn't compile because l is non-copyable 
                                // due to unique_ptr member
higherOrderFunction([p = std::move(p)] {}); // this still works, the closure object is moved

If you pass by const reference, then you cannot pass a mutable lambda that modifies its data members as the argument to higherOrderFunction() because a mutable lambda has a non-const operator(), and you cannot invoke that on a const object.

template <class Function>
void higherOrderFunction(Function const& f);

int i = 0;
higherOrderFunction([=]() mutable { i = 0; }); // will not compile

The best option is to use a forwarding reference. Then higherOrderFunction can accept either lvalues or rvalues that the caller passes.

template <class Function>
void higherOrderFunction(Function&& f) {
     std::forward<Function>(f)();
}

This allows the simple cases as well as the ones mentioned above to compile. For a discussion of why std::forward should be used, see this answer.

Live demo

Answer 2

A copy is a copy, so you cannot mutate the original, might have some performance impact when a lot of data is involved:

#include <iostream>
using namespace std;

template<typename Fn>
void call_value(Fn f) { f(); }
template<typename Fn>
void call_ref(Fn & f) { f(); }
template<typename Fn>
void call_cref(Fn const & f) { f(); }

struct Data {
  Data() {}
  Data(Data const &) {
    cout << "copy" << endl;
  }
  Data(Data &&) {
    cout << "move" << endl;
  }
};

int main(int, char **) {
  Data data;

  auto capref = [&data] () {};
  cout << "capture by value, so we get a ";
  auto capcp = [data] () {};

  cout << " the lambda with a reference ... ";
  call_value(capref);
  cout << " could now be called and mutate .. ";
  call_ref(capref);
  call_cref(capref);
  cout << " but won't, as it had to be declared mutable " << endl;

  cout << "the lambda with an instance: ";
  call_value(capcp);
  cout << "but not ";
  call_ref(capcp);
  call_cref(capcp);
  cout << " the reference versions " << endl;

  bool en = false;
  auto trigger = [en](bool enable = true) mutable {
    if (en) {
      cout << "fire!" << endl;
    }
    if (en or enable) {
      en = true;
    }
  };
  cout << "won't shoot" << endl;
  trigger(false);
  call_value(trigger);
  trigger(false);
  call_ref(trigger);
  cout << "and now ... ";
  trigger(false);
  // const ref won't work
  return 0;
}

See it in action.

Don't forget: lambdas are mere syntactic sugar for callable classes. (But extremely helpful)