得到升压多指标的迭代数字索引 [英] Get numeric index from Boost multi-index iterator

问题描述

我存储了一堆以下

struct Article {
    std::string title;
    unsigned db_id;     // id field in MediaWiki database dump
};

在Boost.MultiIndex的容器，定义为

in a Boost.MultiIndex container, defined as

typedef boost::multi_index_container<
    Article,
    indexed_by<
        random_access<>,
        hashed_unique<tag<by_db_id>,
                      member<Article, unsigned, &Article::db_id> >,
        hashed_unique<tag<by_title>,
                      member<Article, std::string, &Article::title> >
    >
> ArticleSet;

现在我有两个迭代器，一个来自指数＆LT; by_title＆GT; 和一个从指数＆LT; by_id＆GT; 。这是对这些变换索引到容器的随机接入部分的最简单的方法，而无需增加数据成员结构条

Now I've got two iterators, one from index<by_title> and one from index<by_id>. What is the easiest way to transform these to indexes into the random access part of the container, without adding a data member to struct Article?

推荐答案

各项指标利用价值支持新一代的迭代器的<一个href=\"http://www.boost.org/doc/libs/1_44_0/libs/multi_index/doc/reference/ord_indices.html#synopsis\">iterator_to.如果你已经有一个迭代的目标值在一个索引中，你可以使用它来转换为迭代器在另一个索引。

Every index supports generation of an iterator by value using iterator_to. If you already have an iterator to the target value in one index, you could use this to convert to an iterator in another index.

iterator       iterator_to(const value_type& x);
const_iterator iterator_to(const value_type& x)const;

有关转换为指数则可能可以遵循模型 random_access_index.hpp ：

For conversion to index you can likely follow the model in random_access_index.hpp:

  iterator erase(iterator first,iterator last)
  {
    BOOST_MULTI_INDEX_CHECK_VALID_ITERATOR(first);
    BOOST_MULTI_INDEX_CHECK_VALID_ITERATOR(last);
    BOOST_MULTI_INDEX_CHECK_IS_OWNER(first,*this);
    BOOST_MULTI_INDEX_CHECK_IS_OWNER(last,*this);
    BOOST_MULTI_INDEX_CHECK_VALID_RANGE(first,last);
    BOOST_MULTI_INDEX_RND_INDEX_CHECK_INVARIANT;
    difference_type n=last-first;
    relocate(end(),first,last);
    while(n--)pop_back();
    return last;
  }

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