我可以这样创建一个新地址吗？

Question

我正忙着做一个项目。所以我在另一个结构中有一个结构数组。在这种情况下，例如牛中的CattleHealth数组。在我的RecordHealth函数中，我使用牛结构的地址并添加当前列表的当前长度来创建唯一的地址。这样做对吗？

  contract WagyuRecordContract
    {
        address owner;

        struct Cattle
        {
            address RFID;
            string Name;
            uint256 Weight;
            string Gender;
            string Colour;
            string Breed;
            uint Age; 
            uint DOB;
            string Location;
            bool Parent;
            string SireName;
            string DamName;
            bool Active;
            bool ForSale;
            CattleHealth[] HealthRecord;
            CattleGrowth[] GrowthRecord;
            CattleMovements[] MovementsRecord;
            Facility SlaughterDetails;
            Meat[] DistributionDetails
        }


        struct CattleGrowth
        {
            uint DateRecorded;
            uint256 FoodIntake;
            uint256 Growth;
        }


        mapping (address => Cattle) public cattle;
        mapping (address=> CattleGrowth) public growth;           

        modifier Owner() 
        {
            require(msg.sender == owner);
            _;
        }


        function RecordHealth(address rfid, string _bodyCond, string _healthStat, uint256, string _med) Owner public
        {
            health[rfid+cattle[rfid].HealthRecord.length].DateRecorded = now;
            health[rfid+cattle[rfid].HealthRecord.length].BodyCondition = _bodyCond;
            health[rfid+cattle[rfid].HealthRecord.length].HealthStatus = _healthStat;
            health[rfid+cattle[rfid].HealthRecord.length].Medication = _med;
            cattle[rfid].HealthRecord.push(health[rfid+cattle[rfid].HealthRecord.length]);
        }
}

编辑:我的伙伴在没有使用映射的情况下使用了下面的方法。我不明白这是怎么回事。-1是干什么用的？它为什么对牛来说是独一无二的呢？

   function RecordHealth(address rfid, string _bodyCond, string _healthStat, string _med) Owner public
    {
       cattle[rfid].HealthRecord.push(now, _bodyCond, _healthStat, _med)-1;
    }

Answer 1

不是。虽然在C语言或类似语言中指针算法是正常的规则，但在稳健性中，它不能保证工作，无论是对于最终的包装和最终的循环不对齐。因此，强烈建议不要这样做。