是否有一个平面未排序的地图/集合实现? [英] Is there a flat unsorted map/set implementation?

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问题描述

boost.container flat_map和其他元素,Loki AssocVector和许多其他类似元素使元素保持排序.

There is the boost.container flat_map and others, and the Loki AssocVector and many others like these which keep the elements sorted.

是否有一种现代的(实现c ++ 11移动功能等)实现的未排序矢量,适合作为地图/集合?

Is there a modern (c++11 move-enabled, etc.) implementation of an unsorted vector adapted as a map/set?

这个想法是将其用于非常小的地图/集合(少于20个元素)和简单的键(对于这些键,散列并不总是很有意义)

The idea is to use it for very small maps/sets (less than 20 elements) and with simple keys (for which hashing wouldn't always make sense)

推荐答案

像这样吗?

template<class Key, class Value, template<class...>class Storage=std::vector>
struct flat_map {
  struct kv {
    Key k;
    Value v;
    template<class K, class V>
    kv( K&& kin, V&& vin ):k(std::forward<K>(kin)), v(std::forward<V>(vin)){}
  };
  using storage_t = Storage<kv>;
  storage_t storage;

  // TODO: adl upgrade
  using iterator=decltype(std::begin(std::declval<storage_t&>()));
  using const_iterator=decltype(std::begin(std::declval<const storage_t&>()));
  // boilerplate:
  iterator begin() {
    using std::begin;
    return begin(storage);
  }
  const_iterator begin() const {
    using std::begin;
    return begin(storage);
  }
  const_iterator cbegin() const {
    using std::begin;
    return begin(storage);
  }
  iterator end() {
    using std::end;
    return end(storage);
  }
  const_iterator end() const {
    using std::end;
    return end(storage);
  }
  const_iterator cend() const {
    using std::end;
    return end(storage);
  }
  size_t size() const {
    return storage.size();
  }
  bool empty() const {
    return storage.empty();
  }
  // these only have to be valid if called:
  void reserve(size_t n) {
    storage.reserve(n);
  }
  size_t capacity() const {
    return storage.capacity();
  }
  // map-like interface:
  // TODO: SFINAE check for type of key
  template<class K>
  Value& operator[](K&& k){
    auto it = find(k);
    if (it != end()) return it->v;
    storage.emplace_back( std::forward<K>(k), Value{} );
    return storage.back().v;
  }
private: // C++14, but you can just inject the lambda at point of use in 11:
  template<class K>
  auto key_match( K& k ) {
    return [&k](kv const& kv){
      return kv.k == k;
    };
  }
public:
  template<class K>
  iterator find(K&& k) {
    return std::find_if( begin(), end(), key_match(k) );
  }
  template<class K>
  const_iterator find(K&& k) const {
    return const_cast<flat_map*>(this)->find(k);
  }
  // iterator-less query functions:
  template<class K>
  Value* get(K&& k) {
    auto it = find(std::forward<K>(k));
    if (it==end()) return nullptr;
    return std::addressof(it->v);
  }
  template<class K>
  Value const* get(K&& k) const {
    return const_cast<flat_map*>(this)->get(std::forward<K>(k));
  }
  // key-based erase: (SFINAE should be is_comparible, but that doesn't exist)
  template<class K, class=std::enable_if_t<std::is_converible<K, Key>{}>>
  bool erase(K&& k) {
    auto it = std::remove(
      storage.begin(), storage.end(), key_match(std::forward<K>(k))
    );
    if (it == storage.end()) return false;
    storage.erase( it, storage.end() );
    return true;
  }
  // classic erase, for iterating:
  iterator erase(const_iterator it) {
    return storage.erase(it);
  }
  template<class K2, class V2,
    class=std::enable_if_t<
      std::is_convertible< K2, Key >{}&&
      std::is_convertible< V2, Value >{}
    >
  >
  void set( K2&& kin, V2&& vin ) {
    auto it = find(kin);
    if (it != end()){
      it->second = std::forward<V2>(vin);
      return;
    } else {
      storage.emplace_back( std::forward<K2>(kin), std::forward<V2>(vin) );
    }
  }
};

我将容器类型保留为模板参数,因此您可以选择使用类似SBO矢量的结构.

I left the container type as a template argument, so you can use a SBO vector-like structure if you choose.

从理论上讲,我应该公开一个模板参数来替换键上的等号.但是,我确实使键搜索功能透明了.

In theory, I should expose a template parameter for replacing equals on the keys. I did, however, make the key-search functions transparent.

这篇关于是否有一个平面未排序的地图/集合实现?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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