c++ - 从双端队列中删除某个位置的对象

Question

我有这个代码:

 bool tuple_compare(boost::tuple<ppa::Node*, ppa::Node*, ppa::Node*, bool> &tuple_from_done)
  {
  for(int i = 0; i < deque_wait.size(); i++) {

      boost::tuple<ppa::Node*, ppa::Node*, ppa::Node*, bool> tuple_from_wait = deque_wait.at(i);
      ppa::Node *father = boost::get<0>(tuple_from_wait);
      ppa::Node *son = boost::get<0>(tuple_from_wait);
      ppa::Node *second_son = boost::get<2>(tuple_from_wait);

      bool has_seq = boost::get<3>(tuple_from_wait);

      cout << "checking this two " << boost::get<1>(tuple_from_wait)->get_name() <<  " bool sequence "
              <<  boost::get<1>(tuple_from_wait)->node_has_sequence_object  << " and this " 
              << boost::get<2>(tuple_from_wait)->get_name() << " bool seq " <<  boost::get<2>(tuple_from_wait)->node_has_sequence_object
              << " with " << boost::get<0>(tuple_from_done)->get_name() << endl;

      if(boost::get<0>(tuple_from_done)->get_name() == boost::get<1>(tuple_from_wait)->get_name()
              || boost::get<0>(tuple_from_done)->get_name() == boost::get<2>(tuple_from_wait)->get_name())
      {
         cout << " found in here this we need to check if there is something if the sons have a sequences!!!! " << endl; 

         if(boost::get<1>(tuple_from_wait)->node_has_sequence_object == true && boost::get<2>(tuple_from_wait)->node_has_sequence_object == true) 
         {
             cout << " ding, ding, we have one ready!!!" << endl;

             return true;
         }
         else
         {
             cout << "not ready yet" << endl;
         }

        }    

       }

  return false;

}

现在我需要删除在“ding, ding”行中找到的对象，但我不知道该怎么做，我知道迭代器使用得很好我实际上必须从 deque_wait 中删除这个元组并将其移动到 deque_run，但我还不太了解这些，所以你能帮我吗，谢谢。

Answer 1

deque_wait.erase(deque_wait.begin() + i);
//               ^^^^^^^^^^^^^^^^^^^^^^
//               that's an iterator

deque 支持随机访问迭代器，它很像指针(事实上，指针是随机访问迭代器的一种)，所以你可以只获取开始迭代器并向其添加一个整数获取偏移量，就像使用指针一样。