【问题标题】:Upper Bound in Boost Multi-IndexBoost Multi-Index 中的上限
【发布时间】:2014-06-06 16:27:48
【问题描述】:

我正在尝试学习如何使用 Boost 的多索引,并且我一直在专门查看 http://www.boost.org/doc/libs/1_39_0/libs/multi_index/example/basic.cpp 的基本示例。我如何修改它以仅返回年龄小于特定数字(或 id 等)的员工?

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;
}

【问题讨论】:

    标签: c++ boost multi-index


    【解决方案1】:

    您可以在要查询的有序索引上使用上界(或下界,具体取决于您的要求):

    auto const& by_age = get<age>(es);
    auto upper = by_age.upper_bound(32);
    

    比如下面的循环

    for (auto& e : boost::make_iterator_range(by_age.begin(), upper))
        std::cout << e.name << " is " << e.age << " years old\n";
    

    打印

    Albert is 20 years old
    Robert is 27 years old
    Joe is 31 years old
    

    Live On Coliru


    完整代码供参考:

    #include <boost/multi_index_container.hpp>
    #include <boost/multi_index/member.hpp>
    #include <boost/multi_index/ordered_index.hpp>
    #include <boost/range/iterator_range.hpp>
    #include <iostream>
    
    using boost::multi_index_container;
    using namespace boost::multi_index;
    
    struct employee {
        int         id;
        std::string name;
        int         age;
    
        friend std::ostream& operator<<(std::ostream& os,const employee& e) {
            return os << e.id << " " << e.name << " " << e.age << std::endl;
        }
    };
    
    typedef multi_index_container<
      employee,
      indexed_by<
        ordered_unique<tag<struct id>      , BOOST_MULTI_INDEX_MEMBER(employee, int        , id)>   ,
        ordered_non_unique<tag<struct name>, BOOST_MULTI_INDEX_MEMBER(employee, std::string, name)> ,
        ordered_non_unique<tag<struct age> , BOOST_MULTI_INDEX_MEMBER(employee, int        , age)> >
    > employee_set;
    
    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 });
    
        auto const& by_age = get<age>(es);
        auto upper = by_age.upper_bound(32);
    
        for (auto& e : boost::make_iterator_range(by_age.begin(), upper))
            std::cout << e.name << " is " << e.age << " years old\n";
    }
    

    【讨论】:

    • 为什么需要将 by_age 定义为 auto?
    • 你没有。你为什么这么认为?我认为它比boost::multi_index::index&lt;MultiIndexContainer, age&gt;::type const&amp; by_age = get&lt;age&gt;(es) 更易于维护。在这种情况下,"spelling the type out" 甚至不会添加任何文档值,因为类型只是一个类型函数表达式,因此与auto
    • 我明白了,这很有意义!您是如何确定这种类型的?
    • @user3707153 这不是类型;它是一个编译时类型表达式,评估为实际的ordered_index 类型。可以使用 typeid(T).name() and c++filt 或使用编译器对其自身查看真实类型:struct {} _ = get&lt;age&gt;(es);
    • 感谢大家的帮助!
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