嘗試使用字符串數組從類創建新對象

Question

我有一個名為Player的類，其中有三個變量，其中一個是帶有十個值的字符串數組（playerInv）。 如何使用此Player類在我的主活動類中創建一個新的Player，數組中的每個索引都具有空值（“”）？

玩家類別：

public class Player {

    private int playerPos;
    private int playerHP;
    private String playerInv[];

    Player(int startPos, int startHP, String[] newInventory[]) {
        playerPos = startPos;
        playerHP = startHP;
        playerInv = newInventory[10];
    }

    public int getPlayerPos() {
        return playerPos;
    }

    public void setPlayerPos(int playerPos) {
        this.playerPos = playerPos;
    }

    public int getPlayerHP() {
        return playerHP;
    }

    public void setPlayerHP(int playerHP) {
        this.playerHP = playerHP;
    }

    public String setPlayerInv(int pos) {
        return playerInv[pos];
    }

    public void setPlayerInv(String inventory) {
        for (int i = 0; i < 10; i++) {
            this.playerInv[i] = playerInv[i];
        }
    }
}

在主要活動中初始化播放器

public void setupPlayer()
{
    thePlayer = new Player(0,100,);
}

謝謝！

Answer 1

首先，您的類構造函數中存在錯誤。 如果要獲取String數組作為構造函數參數，它應該看起來像這樣：

Player(int startPos, int startHP, String[] newInventory)
{
    playerPos = startPos;
    playerHP = startHP;
    playerInv = newInventory;
}

如果希望playerInv始終是大小為10的String數組，則它應該看起來像這樣。

Player(int startPos, int startHP)
{
    playerPos = startPos;
    playerHP = startHP;
    playerInv = new String[10];
}

然后，使用以下簡單方法創建一個新的Player對象：

thePlayer = new Player(0, 100);

重要的是，在第二種情況下，playerInv變量將包含對填充為null的String類型數組的引用。 因此表將如下所示：

[null, null, null, null, null, null, null, null, null, null]

如果您希望用空字符串（“”）填充數組，如下所示：

   ["","","","","","","","","",""]

然后，您還有至少兩種方法可以執行此操作：

Player(int startPos, int startHP)
{
    playerPos = startPos;
    playerHP = startHP;
    playerInv = new String[10];
    Arrays.fill(playerInv, "");
}

或者，您可以使用以下構造函數：

Player(int startPos, int startHP, String[] newInventory)
{
    playerPos = startPos;
    playerHP = startHP;
    playerInv = newInventory;
}

並在創建新的Player對象時執行以下操作：

String[] inventory = new String[10];
Arrays.fill(inventory, "");
Player thePlayer = new Player(0, 100, inventory);

Answer 2

在您的Player構造函數中：

Player(int startPos, int startHP, String[] newInventory)
{
    playerPos = startPos;
    playerHP = startHP;

    if (newInventory == null)
    {
        playerInv = new String[10];
        for (int i = 0; i < playerInv.length; i++)
        {
            playerInv[i] = "";
        }
    }
    else
    {
        playerInv = newInventory;
    }
}

然后致電：

thePlayer = new Player(0,100,null);

希望能幫助到你...

Answer 3

使用默認情況下滿足您要求的構造函數。