Question

我认为那些是细菌。

我创建了3个这样的对象：

    for (int j = 1; j < 4; j++) {
        int parkingSlot= 1 + rd.nextInt(3);
        AircraftCarrier ac= new AircraftCarrier (fc, j, parkingSlots, parkingSlots);

    }

基于AircraftCarrier类（它的构造函数）：

public AircraftCarrier (FlightControl fc, int idC, int parkingSlots, int freeParkingSlots) {
    this.kontrolaLotow = fc;
    this.id = idC;
    this.ps = parkingSlots;
    this.fps = freeParkingSlots;        
}

所以我有3架航空母舰，对吗？假设我需要为id = 2的载波更改freeParkingSLots值。我该怎么做？

Answer 1

您创建了三个实例，但由于您没有保留对其中任何实例的引用，因此您不再拥有它们。他们被送去垃圾收集。

您需要将每个实例存储在某个集合中以供以后访问。

 List<AircraftCarrier> myList = new ArrayList<AircraftCarrier>();
 for (int j = 1; j < 4; j++) {
    int parkingSlot= 1 + rd.nextInt(3);
    AircraftCarrier ac= new AircraftCarrier (fc, j, parkingSlots, parkingSlots);
    myList.add(ac);
}

Answer 2

您必须将运营商存储在一个数组中：

AircraftCarrier[] carriers = new AircreaftCarrier[3];
for (int j = 0; j < carriers.length; j++) {
    int parkingSlot= 1 + rd.nextInt(3);
    AircraftCarrier ac = new AircraftCarrier (fc, j + 1, parkingSlots, parkingSlots);
    carriers[i] = ac;
}

现在您可以访问它们了：

carriers[1].fps = 6; // You wanted id=2. Since we count from zero in Java, use 1

Answer 3

您创建了每个实例，但无处保存。因此在循环之后，对象“消失”了。使用

List<AircraftCarrier> carriers = new ArrayList<AircraftCarrier>();
for (int j = 1; j < 4; j++) {
  int parkingSlot = 1 + rd.nextInt(3);
  AircraftCarrier ac = new AircraftCarrier (fc, j, parkingSlots, parkingSlots);
  carriers.add(ac);
}

由于您已将id 2分配给第二个元素，因此您现在可以使用carriers.get(1)访问它（get(0)将为您提供第一个元素。）

Answer 4

您必须将运营商“放置”到某处，以便以后可以访问它们：

Map<Integer, AircraftCarrier> carriers = new HashMap<Integer, AircraftCarrier>();

for (int j = 1; j < 4; j++) {
    int parkingSlot= 1 + rd.nextInt(3);
    AircraftCarrier ac= new AircraftCarrier (fc, j, parkingSlots, parkingSlots);
    carriers.put(j, ac);
}

/* update carrier with ID 2 */
carriers.get(2).fps = 1;

设置指定对象的值

4 个答案: