C++:嵌套映射
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
butsdereferences",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
butsdereferences', e.g. consider how we could do a quick & dirty implementation of thatoperator<<
:
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
}
}
}
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