本文介绍了地图中的其他键类型的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
对于特定要求,我想拥有一个具有不同类型键的映射.类似于boost:any. (我有一个旧的gcc版本)
for a particular requirement I want to have a map with keys in different type. Similar to boost:any. (I have an old gcc version)
map<any_type,string> aMap;
//in runtime :
aMap[1] = "aaa";
aMap["myKey"] = "bbb";
使用boost可以做到这一点吗?
is this something possible with the use boost ?
先谢谢
推荐答案
如果您不愿意使用boost变体,则可以破解自己的密钥类型.
If you're not willing to use boost variant, you can hack your own key type.
您可以使用有区别的联合,也可以采用低级技术,只需使用一对std::string
和int:
You could use a discriminated union, or go low-tech en simply use a pair of std::string
and int:
#include <map>
#include <tuple>
#include <iostream>
struct Key : std::pair<int, std::string> {
using base = std::pair<int, std::string>;
Key(int i) : base(i, "") {}
Key(char const* s) : base(0, s) {}
Key(std::string const& s) : base(0, s) {}
operator int() const { return base::first; }
operator std::string() const { return base::second; }
friend bool operator< (Key const& a, Key const& b) { return std::tie(a.first, a.second) < std::tie(b.first, b.second); }
friend bool operator==(Key const& a, Key const& b) { return std::tie(a.first, a.second) == std::tie(b.first, b.second); }
friend std::ostream& operator<<(std::ostream& os, Key const& k) {
return os << "(" << k.first << ",'" << k.second << "')";
}
};
using Map = std::map<Key, std::string>;
int main()
{
Map m;
m[1] = "aaa";
m["myKey"] = "bbb";
for (auto& pair : m)
std::cout << pair.first << " -> " << pair.second << "\n";
}
打印:
(0,'myKey') -> bbb
(1,'') -> aaa
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