如何从multi_index_container获取倒数第二个元素

Question

I have a boost::multi_index_container indexed by hashed_unique and sequenced . How can I get the second from the last element from this container?

struct MyContainer : public mi::multi_index_container<
    MyStruct,
    mi::indexed_by<
        mi::hashed_unique<
          mi::tag<hashed>,
          %some stuff%,
          %some stuff%,
          %some stuff%>
        >,
        mi::sequenced<mi::tag<sequenced> >
    >
>
{ };

As the container is hashed, I can find any element by its hash. But in my case, I do not know the hash of the second-to-last element. However, I know the hash of the last element and hence can get the last element.

MyContainer::iterator myIter = m_table.find(hashOfLast);

Can I use this myIter to get an iterator to the previous element?

Edit:

Can I do something like this?

MyContainer::nth_index<1>::type& seqIdx = m_table.get<1>();
auto current = seqIdx.rbegin();
auto last = seqIdx.rend();

if(current != last){
    current++;
    //How to get the hash of this element now?
}

Answer 1

You can use iterator projection as follows:

MyContainer::index<sequenced>::type::iterator it=
  m_table.get<sequenced>().end(); // iterator to end of sequenced index
--it;--it; // two steps back
MyContainer::iterator myIter=m_table.project<hashed>(it); // project into the hashed index

Note that the same technique can be used for the one-to-last position, which might dispense you with the need to keep your hashOfLast variable.

Can I use this myIter to get an iterator to the previous element?

No (unless you resort to iterator projection as shown above), because of two reasons:

• Hashed index iterators (unlike those of sequenced indices) are not bidirectional (incrementable and decrementable), just forward (incrementable).
• Even if myIter coud be decremented, it wouldn't point to the element at the second-to-last position in the sequenced index: traversal orders in both indices are completely unrelated.