Question

我无法通过从父矢量点到达相关的矢量来获取矢量的大小。我已经验证调用函数myfunc 1 2 3 4 5会创建五个Person对象并将它们放在left向量中。但是当我尝试通过从bridge到left的点走时返回相同的大小时，我得到0作为大小。

我做错了什么？

int main(int argc, char* argv[]) {
    Person* p_ptr;
    int id_source = 0;
    vector<Person> left;
    vector<Person> right;
    bridge.push_back(left);
    bridge.push_back(right);
    cout << "bridge.size() = " << bridge.size() << endl;
    for (int i = 1; i < argc; i++) {
      id_source++;
      cout << "Creating Person with crossing speed of " << argv[i] << " and id of " << id_source << endl;
      p_ptr = new Person(atoi(argv[i]), id_source);
      left.push_back(*p_ptr);
    }
    /*SIZE TESTING*/
    cout << "Left side of bridge has " << left.size() << " people on it " << endl;
    cout << "bridge.at(0).size() = " << bridge.at(0).size() << endl;
    cout << "bridge.at(1).size() = " << bridge.at(1).size() << endl;
    int slowest_id = get_slowest(0);
    for (int i = 0; i < left.size(); i++) {
      if (slowest_id == left.at(i).get_id()) {
        p_ptr = &left.at(i);
      }
    }
    cout << "The slowest person has id of " << slowest_id << " and speed of " << p_ptr->get_crossing_time() << endl;
  }
}

Answer 1

left和bridge[0]是两个不同的列表。当您致电bridge.push_back(left)时，您会复制当前的left列表（空白）。稍后添加的元素将不在bridge版本中。

如何获得另一个向量内的向量大小？

1 个答案: