Why the [] operator in sorted_vector_map?

Question

I do not understand what the operator [] does in sorted_vector_map.

Specifically when the key does not exist, what value is added to the data structure?
What is value_type(key, mapped_type())?
Is it a constructor call to std::pair by default?
What is mapped_type()?
Is it also a constructor call?

mapped_type& operator[](const key_type& key) {
    iterator it = lower_bound(key);
    if (it == end() || key_comp()(key, it->first)) {
      return insert(it, value_type(key, mapped_type()))->second;
    }
    return it->second;
}

Code is from the following link...

https://github.com/facebook/folly/blob/master/folly/sorted_vector_types.h#L1097

score 2 · Answer 1 · answered Dec 27 '20 at 19:59

The answers are in the header file.

Consider:

value_type(key, mapped_type())

One line 743 of the file you linked, you'll see this declaration:

typedef typename Container::value_type value_type;

But what's Container? On line 728 you'll find that Container is a template argument, which is probably a std::pair (unless the user has supplied another one).

class Container = std::vector<std::pair<Key, Value>, Allocator>>

So yes, that line is a constructor call to initialize a std::pair, because that's what this particular data structure uses as its value.

mapped_type() is also a constructor call, with no arguments. It's similar to:

int i = int();

I get that Container is std::pair, but why is container::value_type also std::pair? Is this by design? — user1559897, Dec 27 '20 at 20:05
No `Container` is not `std::pair` it is - as you wrote - `std::vector, Allocator>>` (`std::vector`). — t.niese, Dec 27 '20 at 20:08

t.niese · Accepted Answer · 2020-12-27T20:29:08.440

Container is the template argument that defines what container is used by sorted_vector_map to store the key-value pairs and defaults to a std::vector(std::vector<std::pair<Key, Value>, Allocator>>)

value_type is Container::value_type (typedef typename Container::value_type value_type;) which (for the default template argument) is std::pair<Key, Value> (see std::vector Member types)

mapped_type is Value (typedef Value mapped_type;) so the type of the value stored in the sorted_vector_map

What is value_type(key, mapped_type())?
What is mapped_type()?
Is it also a constructor call?

So value_type(key, mapped_type()) creates a std::pair with key as first, and a default constructed Value (mapped_type()) as second.

Is it a constructor call to std::pair by default?

yes

template <
    class Key,
    class Value, // <<===============
    class Compare = std::less<Key>,
    class Allocator = std::allocator<std::pair<Key, Value>>,
    class GrowthPolicy = void,
    class Container = std::vector<std::pair<Key, Value>, Allocator>>   // <<===============
class sorted_vector_map : detail::growth_policy_wrapper<GrowthPolicy> {
  detail::growth_policy_wrapper<GrowthPolicy>& get_growth_policy() {
    return *this;
  }

  template <typename K, typename V, typename C = Compare>
  using if_is_transparent =
      _t<detail::sorted_vector_enable_if_is_transparent<void, C, K, V>>;

  struct EBO;

 public:
  typedef Key key_type;
  typedef Value mapped_type; // <<===============
  typedef typename Container::value_type value_type; // <<===============
  typedef Compare key_compare;
  typedef Allocator allocator_type;
  typedef Container container_type;

If i had a sorted_vector_container object called sv and I were to create an object of type mapped_type, can I do sv:: mapped_type() to invoke the constructor? — user1559897, Dec 27 '20 at 22:15
@user1559897 don't ask new questions in the comment. If you have a new question create a new one. You have to use either `decltype` to get the type of `sv` (`decltype(test)::mapped_type`) or use `sorted_vector_container::mapped_type` with `Key` and `Value` being the types you used. — t.niese, Dec 28 '20 at 08:39

Why the [] operator in sorted_vector_map?

2 Answers2