c++ - 从 Vector 生成类数据成员

Question

请考虑这个 C++ 问题:

#include <cstdio>
#include <iostream>
#include <vector>
using namespace std;

class ParseURL{
    private:
        int qualify; // 0 means we qualify
    public:
        string node;
        string service;
        bool primary;
        bool health;
        string epoch;
        // methods
        ParseURL(string URL);
        ~ParseURL();
};


ParseURL::ParseURL(string URL){
    vector<string> g2 = {"node", "service", "primary", "health", "epoch"};

    for (string tag : g2){
        auto found = URL.find(tag);
        if ( found != string::npos ){
            auto cut_from = found + 1 + tag.size() ;
            auto meh = URL.substr(cut_from, URL.substr(cut_from).find("&") );
            // is there a way we can avoid these lines below?
            if (tag.find("node",0) == 0){
                this->node = meh;
            } else if (tag.find("service",0) == 0 ){
                this->service = meh;
            } else if (tag.find("epoch",0) == 0) {
                this->epoch = meh;
            } else if (tag.find("health",0) == 0){
                this->health = (meh == "OK");
            } else if (tag.find("primary",0) == 0 ){
                this->primary == (this->node == meh);
            }
        }
    }

}

ParseURL::~ParseURL(){
    cout << "Tearing Down the class\n";
}
int main(){
    char req[] = "GET /register?node=hostrpi3&service=potatoservice&primary=node1&health=OK&epoch=1559345106 HTTP";
    string Request = req;
    Request = Request.substr(Request.find_first_of(" ") );
    Request = Request.substr(0, Request.find(" HTTP"));
    ParseURL *a = new ParseURL(Request);

    cout.width(12);
    cout << "a->node: " << a->node << endl;
    cout.width(12);
    cout << "a->service: " << a->service << endl;
    cout.width(12);
    cout << "a->epoch: " << a->epoch << endl;
    cout.width(12);
    cout << "a->health: " << a->health << endl;
    cout.width(12);
    cout << "a->primary: " << a->primary << endl;

    delete(a);

    return 0;
}

我需要表示一个基于 URL 查询字符串的对象，我需要的标签在 vector g2 中表示，如您所见，我从这些标签中为 ParseURL 创建了数据成员。

O/P 看起来像:

   a->node: hostrpi3
a->service: potatoservice
  a->epoch: 1559345106
 a->health: 1
a->primary: 0
Tearing Down the class

虽然这个版本可以，但感觉可以改进很多。尽管 Vector 中包含了所有内容，但我仍在为每个属性重复执行 tag.find。我希望一定有更好的方法。你能指出正确的方向吗？谢谢!

编辑 1

谢谢，我按照建议更新了构造函数:

ParseURL::ParseURL(string URL){
    char tags[10][100] = {"node", "service", "primary", "health", "epoch"};

    int i = 0;
    for (auto tag : tags){
        auto found = URL.find(tag);
        if ( found != string::npos ){
            auto cut_from = found + 1 + strlen(tag);
            auto tag_info = URL.substr(cut_from, URL.substr(cut_from).find("&") );
            switch (i){
                case 0:
                    this->node = tag_info; 
                    break;
                case 1:
                    this->service = tag_info; 
                    break;
                case 2:
                    this->primary = (this->node == tag_info); 
                    break;
                case 3:
                    this->health = (tag_info == "OK"); 
                    break;
                case 4:
                    this->epoch = tag_info; 
                    break;
            }
        }
    i++;
    }
}

例如，bash 提供变量的间接引用——也就是说，一个变量可能包含另一个变量的名称:

$ cat indirect.sh 
#!/bin/bash 

values="value1 value2"
value1="this is value1"
value2="this is value2"

for i in $values; do 
    echo "$i has value ${!i}"
done

$ ./indirect.sh 
value1 has value this is value1
value2 has value this is value2

我希望像这样的案例也有类似的东西，不过，我在这里学到了宝贵的经验教训。谢谢你!

Answer 1

假设您想要保留不同的类成员(而不是一个键值存储)，您可以使用一个包含字符串和成员指针的静态表:

vector<pair<string,string ParseURL::*>> keys={{"node",&ParseURL::node},…};

但这在这里行不通，因为类型不同，而且每种类型的处理也不相同。相反，对于这种几乎并行，语法上的最佳选择通常是lambda:

const auto get=[&URL](const string &tag) {
  auto found = URL.find(tag);
  if(found == string::npos) return "";
  auto cut_from = found + 1 + tag.size() ;
  return URL.substr(cut_from, URL.substr(cut_from).find("&") );
};
node=get("node");
service=get("service");
epoch=get("epoch");
health=get("health")=="OK";
primary=get("primary")==node;

这种风格的好处是可以明确初始化health和primary。通过对私有(private)构造函数进行充分的扭曲，可以将其放入成员初始化列表中，这是更好的做法。