istringstream的操作

今天在*上看到这么个问题,写完之后看了看别人的提交的答案,感觉自己的答案虽然能得出正确结果但是有点啰嗦,对于c++还是没有熟练,没有想起有istringstream,而且提问的老外写的程序bug实在是太多了,while循环啊,分号啊都不规范,不过那个return挺有意思,嘿嘿第一次见(可能是写的代码和看的代码太少了吧 ,嘿嘿 ,见笑,我也是一菜),于是我放弃了我原来的答案,修改了一下别人的答案,那个人的答案也不对,不能运行啊,所以说我自己改动了一下.

今儿记录一下,省的以后忘..

老外的提问:

 struct client
 {
     string PhoneNumber;
     string FirstName;
     string LastName;
     string Age;
 };
 int main()
 {
     string data = getClientDatabase();

     vector <client> clients;

     parse_string(data, clients);
     return ;
 }

 string getClientDatabase()
 {
     return
         "[phone] Sofia Ross 57 \n"
         "[phone] Jenna Doh 30 \n"
         "[phone] Emily Saks 43 \n"

 }

 void parse_string(string data, vector <client> &clients)
 {
     string temp;
     string temp1;
     string temp2;
     string temp3;

     int i = ;
     int j = ;
     int k = ;
     int l = ;

     while (i < data.length())
     {
         if (data.at(i) != ' ')
         {
             temp.push_back(data.at(i));
             j++;
         }
         else
         {
             clients.at(i).PhoneNumber = temp;
         }

     }
     if (data.at(j) != ' ')
     {
         temp1.push_back(data.at(j));
         k++;
     }
     else
     {
         clients.at(i).FirstName = temp1;
     }

     if (data.at(k) != ' ')
     {
         temp2.push_back(data.at(k));
         l++;
     }
     else
     {
         clients.at(i).LastName = temp2;
     }

     if (data.at(l) != ' ')
     {
         temp3.push_back(data.at(l));

     }
     else
     {
         clients.at(i).Age = temp3;
     }
     i++;

 }

我的回答:

 void parse_string(string data, vector <client> & clients)
  {
      struct client tempS;
      istringstream iss(data);
      for (size_t i=; iss >> tempS.PhoneNumber; ++i)
      {
          iss >> tempS.FirstN >> tempS.LastN >> tempS.Age;
          clients.push_back(tempS);
      }
  }
 //test code

 vector <client>::iterator it = clients.begin();
      for(; it != clients.end(); ++it )
     {
         cout << it->PhoneNumber << " "
              << it->FirstN << " "
              << it->LastN << " "
              << it->Age << endl;
     }