如何在Boost Multi-index中使equal_range迭代器按不同索引排序？

Question

我在类上有一个 boost 多索引容器employee（取自 boost 官方文档）：

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;

下面是通过（id，name，age）打印的容器中的数据示例：

0 Joe 31
1 Robert 27
2 John 40
3 Albert 20
4 John 57
5 John 58
6 John 22

我想要一个迭代器，其中包含按年龄（最后一个字段）排序的名称为 John 的所有项目。我尝试了equal_range方法：

auto iter1 = boost::make_iterator_range(es.get<name>().equal_range("John"));

它返回一个迭代器，其中包含名为 John 的所有记录。如何使这个迭代器按第三个索引（即年龄）排序？

输出应该是：

6 John 22
2 John 40
4 John 57
5 John 58

Answer 1

好的。这是再现器Live On Coliru

输出确实是

2 "John" 40
4 "John" 57
5 "John" 58
6 "John" 22

现在，如果还要按年龄排序（注意单词的选择），您可以使用复合键。所以而不是：

    bmi::ordered_non_unique<
        bmi::tag<struct name>,
        bmi::member<employee, std::string, &employee::name>
    >,

使用

    bmi::ordered_non_unique<
        bmi::tag<struct name_age>,
        bmi::composite_key<employee,
            bmi::member<employee, std::string, &employee::name>,
            bmi::member<employee, int, &employee::age>
        >
    >,

现在你可以

for (employee const& emp : boost::make_iterator_range(es.get<name_age>().equal_range("John"))) {
    std::cout << emp.id << " " << std::quoted(emp.name) << " " << emp.age << "\n";
}

印刷

6 "John" 22
2 "John" 40
4 "John" 57
5 "John" 58

完整样本

Live On Coliru

#include <boost/multi_index/composite_key.hpp>
#include <boost/multi_index/member.hpp>
#include <boost/multi_index/ordered_index.hpp>
#include <boost/multi_index_container.hpp>
#include <boost/range/iterator_range.hpp>

namespace bmi = boost::multi_index;

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

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

#include <iostream>
#include <iomanip>

int main() {
    employee_set es {
        {0, "Joe",    31},
        {1, "Robert", 27},
        {2, "John",   40},
        {3, "Albert", 20},
        {4, "John",   57},
        {5, "John",   58},
        {6, "John",   22},
    };

    std::cout << "name index:\n";
    for (employee const& emp : boost::make_iterator_range(es.get<name>().equal_range("John"))) {
        std::cout << emp.id << " " << std::quoted(emp.name) << " " << emp.age << "\n";
    }

    std::cout << "name_age index:\n";
    for (employee const& emp : boost::make_iterator_range(es.get<name_age>().equal_range("John"))) {
        std::cout << emp.id << " " << std::quoted(emp.name) << " " << emp.age << "\n";
    }
}

印刷

name index:
2 "John" 40
4 "John" 57
5 "John" 58
6 "John" 22
name_age index:
6 "John" 22
2 "John" 40
4 "John" 57
5 "John" 58