检查是否为null Bluej(Java)不起作用

时间:2016-04-07 17:21:20

标签: java null bluej

我已经对此进行了广泛的研究,并找到了检查字符串是否为空的答案,但是在检查我实例化的类是否为空时却没有。这是一个由另一个类实例化的类,它保存所有房间的列表,就像Cave Adventure游戏一样。这是代码:

public class Room 
{
    private String description;
    private Room northExit;
    private Room southExit;
    private Room eastExit;
    private Room westExit;

    /**
     * Create a room described "description". Initially, it has
     * no exits. "description" is something like "a kitchen" or
     * "an open court yard".
     * @param description The room's description.
     */
    public Room(String description) 
    {
        this.description = description;
    }

    /**
     * Define the exits of this room.  Every direction either leads
     * to another room or is null (no exit there).
     * @param north The north exit.
     * @param east The east east.
     * @param south The south exit.
     * @param west The west exit.
     */
    public void setExits(Room north, Room east, Room south, Room west) 
    {
        if(north != null)
            northExit = north;
        if(east != null)
            eastExit = east;
        if(south != null)
            southExit = south;
        if(west != null)
            westExit = west;
    }

    /**
     * @return The description of the room.
     */
    public String getDescription()
    {
        return description;
    }

    public Room getExit(String direction)
    {
        if(direction.equals("north")) {
            return northExit;
        }
        if(direction.equals("east")) {
            return eastExit;
        }
        if(direction.equals("south")) {
            return southExit;
        }
        if(direction.equals("west")) {
            return westExit;
        }
            return null;
    }


    public String getExitString() {
        if (!northExit.equals(null) && !northExit.equals("")) 
            return "north ";

        if (!eastExit.equals(null)  && !eastExit.equals("")) 
            return "east ";

        if (!southExit.equals(null)  && !southExit.equals("")) 
            return "south ";

        if (!westExit.equals(null)  && !westExit.equals(""))
            return "west ";

       else {
           System.out.println("There are no doors!");
           return null;
        }

    }
}

当它到达getExitString()方法时,我最终得到一个NullPointerException。

我已经花了很多时间研究这个问题而且我目前处于挫折的极限,任何帮助都会非常感激。

1 个答案:

答案 0 :(得分:1)

表达式northExit.equals(null)(作为一个示例)将无法正常工作。

以这种方式思考:northExit.位意味着northExit引用的解除引用,以找到它指向的内容。如果null null这样的取消引用会给您提供您所看到的例外情况。

检查值是否为if ((northExit != null) && (! northExit.equals(""))) ... 的正确方法是使用引用相等(a),如下所示:

==

(a)在教学时,我经常发现从学生那里借了几张五美元的笔记,并对其进行解释。

就参考平等.equals()而言,它们之间存在差异,因为它们实际上是不同的物理项目。在内容或价值平等abstract class Edge2D[T : Point2DInterface] { val p1: T val p2: T def length(): Double = { implicitly[Point2DInterface[T]].Sub(p1, p2) } } trait Point2DInterface[T] { def Sub(first: T, second: T): Double } 方面,它们是相同的。

然后我掏了十块钱,希望他们在课程结束时忘掉它,这是我收入的一个很好的补充: - )