C++ deque iterate to second last element -

i trying iterate through deque excluding last entry. ideally avoid counting , comparing length, tried

    auto it_end = dq.rbegin(); it_end++;     ( auto = dq.begin(); !=it_end; ) {         if ( cond() ) {             = dq.erase( );         } else {             it++;         }     } 

but compiler complains there no match operand seems understandable since have reverse iterator , regular iterator. there elegant way stop before last element avoids counting? like, offset? tried usign back, turned out reference not iterator, != not happy either.

you can use dq.end() - 1 1 before end.

you can use std::remove or std::remove_if remove items want instead of doing job on own. note kind of partitioning operation--it returns iterator, , want keep beginning of range iterator. want delete after iterator, end of range provided input.

in case, might on general order:

std::deque<int> vals { 1, 2, 3, 4, 5, 6, 7, 8};  // iterator 1 before end of `vals` auto end = vals.end()-1;  // remove items in range (so except `8`) auto pos = std::remove_if(vals.begin(), end,      [](int v){ return v %2 == 0;}  );  // erase items "removed" vals.erase(pos, end);  // show result (v : vals)     std::cout << v << ", ";