What does it mean for a lambda to be static?

Question

Say I have a binary search function which initializes and uses a lambda:

bool custom_binary_search(std::vector<int> const& search_me)
{
  auto comp = [](int const a, int const b)
    {
      return a < b;
    };

  return std::binary_search(search_me.begin(), search_me.end(), comp);
}

Without pointing out that this is completely redundant and just focusing on the lambda; is it expensive to be declaring and defining that lambda object every time? Should it be static? What would it mean for a lambda to be static?

Answer 1

The variable 'comp' with type <some anonymous lambda class> can be made static, pretty much as any other local variable, i.e. it is the same variable, pointing to the same memory address, every time this function is run).

However, beware of using closures, which will lead to subtle bugs (pass by value) or runtime errors (pass-by-reference) since the closure objects are also initialized only once:

bool const custom_binary_search(std::vector<int> const& search_me, int search_value, int max)
{
  static auto comp_only_initialized_the_first_time = [max](int const a, int const b)
  {
      return a < b && b < max;
  };

  auto max2 = max;
  static auto comp_error_after_first_time = [&max2](int const a, int const b)
  {
      return a < b && b < max2;
  };

  bool incorrectAfterFirstCall = std::binary_search(std::begin(search_me), std::end(search_me), search_value, comp_only_initialized_the_first_time);
  bool errorAfterFirstCall = std::binary_search(std::begin(search_me), std::end(search_me), search_value, comp_error_after_first_time);

  return false; // does it really matter at this point ?
}

Note that the 'max' parameter is just there to introduce a variable that you might want to capture in your comparator, and the functionality this "custom_binary_search" implements is probably not very useful.