c++ - 查找所有在 Boost 中排序的

Question

使用 Boost 网站上提供的示例:http://www.boost.org/doc/libs/1_55_0/libs/multi_index/example/basic.cpp我已经学会了如何使用迭代器通过索引获取元素。但是，我想让所有索引都匹配并按另一个索引排序。

例如，在给定的示例中，我想获取所有名为“John”的员工，但返回给我的结果按年龄排序。截至目前，当我得到名字时，我会返回一个迭代器到“约翰”的第一次出现，等等，但是除了插入记录的顺序之外，它们的返回方式没有任何顺序.

请帮忙!

相关代码:

/* Boost.MultiIndex basic example.
 *
 * Copyright 2003-2008 Joaquin M Lopez Munoz.
 * Distributed under the Boost Software License, Version 1.0.
 * (See accompanying file LICENSE_1_0.txt or copy at
 * http://www.boost.org/LICENSE_1_0.txt)
 *
 * See http://www.boost.org/libs/multi_index for library home page.
 */

#if !defined(NDEBUG)
#define BOOST_MULTI_INDEX_ENABLE_INVARIANT_CHECKING
#define BOOST_MULTI_INDEX_ENABLE_SAFE_MODE
#endif

#include <boost/multi_index_container.hpp>
#include <boost/multi_index/member.hpp>
#include <boost/multi_index/ordered_index.hpp>
#include <algorithm>
#include <iostream>
#include <iterator>
#include <string>

using boost::multi_index_container;
using namespace boost::multi_index;

/* an employee record holds its ID, name and age */

struct employee
{
  int         id;
  std::string name;
  int         age;

  employee(int id_,std::string name_,int age_):id(id_),name(name_),age(age_){}

  friend std::ostream& operator<<(std::ostream& os,const employee& e)
  {
    os<<e.id<<" "<<e.name<<" "<<e.age<<std::endl;
    return os;
  }
};

/* tags for accessing the corresponding indices of employee_set */

struct id{};
struct name{};
struct age{};

/* see Compiler specifics: Use of member_offset for info on
 * BOOST_MULTI_INDEX_MEMBER
 */

/* Define a multi_index_container of employees with following indices:
 *   - a unique index sorted by employee::int,
 *   - a non-unique index sorted by employee::name,
 *   - a non-unique index sorted by employee::age.
 */

typedef multi_index_container<
  employee,
  indexed_by<
    ordered_unique<
      tag<id>,  BOOST_MULTI_INDEX_MEMBER(employee,int,id)>,
    ordered_non_unique<
      tag<name>,BOOST_MULTI_INDEX_MEMBER(employee,std::string,name)>,
    ordered_non_unique<
      tag<age>, BOOST_MULTI_INDEX_MEMBER(employee,int,age)> >
> employee_set;

template<typename Tag,typename MultiIndexContainer>
void print_out_by(
 const MultiIndexContainer& s,
 Tag* =0 /* fixes a MSVC++ 6.0 bug with implicit template function parms */
)
{
  /* obtain a reference to the index tagged by Tag */

  const typename boost::multi_index::index<MultiIndexContainer,Tag>::type& i=
    get<Tag>(s);

  typedef typename MultiIndexContainer::value_type value_type;

  /* dump the elements of the index to cout */

  std::copy(i.begin(),i.end(),std::ostream_iterator<value_type>(std::cout));
}


int main()
{
  employee_set es;

  es.insert(employee(0,"Joe",31));
  es.insert(employee(1,"Robert",27));
  es.insert(employee(2,"John",40));

  /* next insertion will fail, as there is an employee with
   * the same ID
   */

  es.insert(employee(2,"Aristotle",2387));

  es.insert(employee(3,"Albert",20));
  es.insert(employee(4,"John",57));

  /* list the employees sorted by ID, name and age */

  std::cout<<"by ID"<<std::endl;
  print_out_by<id>(es);
  std::cout<<std::endl;

  std::cout<<"by name"<<std::endl;
  print_out_by<name>(es);
  std::cout<<std::endl;

  std::cout<<"by age"<<std::endl;
  print_out_by<age>(es);
  std::cout<<std::endl;

  return 0;
}

Answer 1

您可以通过以下两种方式之一进行:

1:将第二个索引替换为基于 composite key 的索引的(姓名，年龄):

typedef multi_index_container<
  employee,
  indexed_by<
    ordered_unique<
      tag<id>,  member<employee,int,&employee::id>>,
    ordered_non_unique<
      tag<name>,
      composite_key<
        employee,
        member<employee,std::string,&employee::name>,
        member<employee,int,&employee::age>
      >
    >,
    ordered_non_unique<
      tag<age>, member<employee,int,&employee::age>>
  >
> employee_set;

int main()
{
  employee_set es={
    {0,"John",31},{1,"Robert",27},{2,"John",57},
    {5,"John",2387},{3,"Albert",20},{4,"John",40}};

  for(auto p=es.get<name>().equal_range("John");p.first!=p.second;++p.first){
    std::cout<<*(p.first);
  }
}

2:对结果进行排序:

template<typename Iterator>
std::vector<std::reference_wrapper<const employee>>
sort_by_age(std::pair<Iterator,Iterator> p)
{
  std::vector<std::reference_wrapper<const employee>> v(p.first,p.second);
  std::sort(v.begin(),v.end(),
    [](const employee& e1,const employee& e2){return e1.age<e2.age;});
  return v;
}

int main()
{
  employee_set es={
    {0,"John",31},{1,"Robert",27},{2,"John",57},
    {5,"John",2387},{3,"Albert",20},{4,"John",40}};

  for(auto e:sort_by_age(es.get<name>().equal_range("John"))){
    std::cout<<e;
  }
}

哪个更好？ #1 在没有进一步后处理的情况下给出了所需的结果，但是插入索引的速度(稍微)慢了，因为比较标准比按名称进行的简单比较更昂贵。 #2 具有后处理惩罚，但可以与年龄以外的排序标准一起使用(对于 #1，第二个键在定义时固定)。