Question

我是c ++的新手，我想知道你是否可以在数组中存储对象名，以便稍后使用for循环访问它们。例如：

SELECT s3.userName, s3.taskName, AVG(s3.failCount) AS avgFailCount
FROM (
    SELECT s1.userName, s1.taskName, s1.ts, s1.PreviousTS, COALESCE(s2.failCount,0) AS failCount
    FROM (
        SELECT t1.userName, t1.taskName, t1.ts, LAG(t1.ts) OVER (PARTITION BY t1.userName, t1.taskName ORDER BY t1.ts) AS PreviousTS --ROW_NUMBER() OVER (PARTITION BY t1.userName ORDER BY t1.ts) AS rn
        FROM #taskevents t1
        WHERE t1.eventName = 'success'
    ) s1
    OUTER APPLY ( 
        SELECT t2.userName, t2.taskName,  COUNT(*) AS failCount 
        FROM #taskevents t2
        WHERE t2.eventName = 'fail'
            AND t2.userName = s1.userName
            AND t2.taskName = s1.taskName
            AND t2.ts < s1.ts  
            AND ( t2.ts >= s1.PreviousTS OR s1.PreviousTS IS NULL )
        GROUP BY t2.userName, t2.taskName
    ) s2
) s3
GROUP BY s3.userName, s3.taskName

Answer 1

我能想到的最简单的C ++方法是使用std::map<std::string,Class>：

#include <iostream>
#include <string>
#include <map>

using namespace std;

class Class {
    public:
        string name; // the name of the object
        int age; // the age of the object
};

int main() {
  std::map<std::string,Class> objrefMap {
      { "Obj1", Class() },
      { "Obj2", Class() } ,
      { "Obj3", Class() }
  };

  objrefMap["Obj1"].name = "Judy";
  objrefMap["Obj1"].age = 29;
  objrefMap["Obj2"].name = "Mike";
  objrefMap["Obj2"].age = 38;
  objrefMap["Obj3"].name = "Dorothy";
  objrefMap["Obj3"].age = 19;

  for(auto objitem : objrefMap) {
      cout << objitem.second.name << endl;
      cout << objitem.second.age << endl;
  }
}

请参阅live demo。

Answer 2

我尝试编辑您的代码以满足您的需求。然而，这不是最佳解决方案，需要付出很多努力。你可以改用地图。希望这很有用。

#include <iostream>
using namespace std;

class Class {
public:
    string name; // the name of the object
    int age; // the age of the object
}; //don't forget the semicolon here.

int main()
{
    int nameCount = 0;
    int ageCount = 0;
    string nameArray[10];
    int ageArray[10];

    Class Obj1;
    Class Obj2;
    Class Obj3;

    nameArray[nameCount] = Obj1.name = "Judy"; nameCount++;
    ageArray[ageCount]  = Obj1.age = 29; ageCount++;
    nameArray[nameCount] = Obj2.name = "Mike"; nameCount++;
    ageArray[ageCount]  = Obj2.age = 38; ageCount++;
    nameArray[nameCount] = Obj3.name = "Dorothy"; nameCount++;
    ageArray[ageCount]  = Obj3.age = 19; ageCount++;

    for(int i = 0, j = 0; i < nameCount && j < ageCount; i++, j++)
    {
        cout << nameArray[i] << endl;
        cout << ageArray[i] << endl;
    }
}

输出：

Judy
29
Mike
38
Dorothy
19

在数组中存储对象名称？

2 个答案: