问题描述

定义如下:

struct nmap;struct nmap: map>{};

下面的最后一行不起作用:

The last line below doesn't work:

nmap my_map;my_map["a"] = "b";my_map["c"] = 新的 nmap;my_map["c"]["d"] = "e";

我需要添加什么才能使其正常工作?

What do I need to add, in order for this to work?

推荐答案

我建议要么选择一个小巧易读的助手:

I'd suggest either going for a tiny little readable helper:

#include <boost/variant.hpp>
#include <map>

using std::map;

struct nmap;
struct nmap: map<std::string, boost::variant<std::string, nmap*>>
{
    typedef boost::variant<std::string, nmap*> Variant;
    typedef map<std::string, Variant> base;

    friend nmap&       as_map(Variant& v)       { return *boost::get<nmap*>(v); }
    friend nmap const& as_map(Variant const& v) { return *boost::get<nmap*>(v); }

    friend std::string&       as_string(Variant& v)       { return boost::get<std::string>(v); }
    friend std::string const& as_string(Variant const& v) { return boost::get<std::string>(v); }
};

int main()
{
    nmap my_map;
    my_map["a"] = "b";
    my_map["c"] =  new nmap;

    as_map(my_map["c"])["d"] = "e";
}

或者采用递归变体方式.让我画草图:

Or go the recursive variant way. Let me sketch:

这让 IMO 更优雅:

This is IMO more elegant:

#include <boost/variant.hpp>
#include <map>

using nmap  = boost::make_recursive_variant<std::string, std::map<std::string, boost::recursive_variant_> >::type;
using map_t = std::map<std::string, nmap>;

#include <iostream>
static std::ostream& operator<<(std::ostream& os, nmap const& map);

int main()
{
    nmap my_map = map_t
    {
        { "a", "b" },
        { "c", map_t
            {
                { "d", "e" },
                { "f", map_t
                    {
                        { "most nested", "leaf node" },
                        { "empty", map_t { } },
                    } },
            } },
        };

    std::cout << my_map;
}

乍一看这可能看起来更复杂，但它实际上有许多重要的优点:

At first glance this may look more complicated, but it actually has a number of important advantages:

• 不再从不打算继承的类继承
• 不再限制根"对象/必须是一个映射(现在它也可以是一个字符串，因此变体更加一致)
• 不再有内存泄漏 由于该变体现在实际上负责分配实际的 nmap 实例，因此具有全部价值 -开箱即用的语义.

• 允许对树进行惯用的访问，不需要ifs and buts dereferences"，例如考虑我们如何做一个快速的&operator<< 的脏实现:

• no more inheriting from classes not intended for inheritance
• no more limitation that the 'root' object /must/ be a map (it can be a string too now, so the variant is more consistently honoured)
• no more memory leaks Due to the fact that the variant now actually takes care of allocating the actual nmap instances, there's full value-semantics out of the box.

• allows for idiomatic visiting of the tree, no need for 'ifs and buts dereferences', e.g. consider how we could do a quick & dirty implementation of that operator<<:

static std::ostream& operator<<(std::ostream& os, nmap const& map)
{
    struct print : boost::static_visitor<void>
    {
        print(std::ostream& os, int indent = 0) : os(os), indent(indent) { }

        void operator()(map_t const& map) const {
            os << "{";
            for(auto& item : map)
            {
                os << "
";
                do_indent();
                os << "    " << item.first << ": ";
                boost::apply_visitor(print(os, indent+4), item.second);
            }
            if (!map.empty()) { os << "
"; do_indent(); };
            os << "}";
        }

        void operator()(std::string const& str) const {
            os << str;
        }

    private:
        std::ostream& os;
        void do_indent() const { for (int n = indent; n>0; --n) os << ' '; }
        int indent = 0;
    };

    boost::apply_visitor(print(os), map);
    return os;
}

看到它在coliru上直播

See it Live On coliru

输出:

# g++ -std=c++11 -Wall -pedantic -Wextra main.cpp  && ./a.out

{
    a: b
    c: {
        d: e
        f: {
            empty: {}
            most nested: leaf node
        }
    }
}

这篇关于C++:嵌套映射的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！