我正在嘗試在我正在編寫的程序中創(chuàng)建兩個玩家。問題是可以創(chuàng)建三種類型的玩家，但為了稍后引用它們，我希望將它們命名為玩家一和玩家二。三個類：“Human”、“Computer”和“Karen”是“Player”的派生類。我想知道如何創(chuàng)建一個數(shù)組 Player，然后我可以將player1 和player2 對象放入其中。這可以嗎？int playerType = 0;int highestScore = 0;Player[] player = new Player[2]; // This is the array of type Player, the master classScanner userInput = new Scanner (System.in);for (int i = 1; i < 3; i++){ System.out.println ("Choose a type of player:"); System.out.println ("1: Human"); System.out.println ("2: Computer (Threshold = 20)"); System.out.println ("3. Computer (Custom)"); System.out.print ("? : "); playerType = Integer.valueOf(userInput.nextLine()); // gets input choice from user switch (playerType) { case 1: Human player[i] = new Human(); case 2: Computer player[i] = new Computer(); case 3: Karen player[i] = new Karen(); }}我嘗試了另一種方法，使用 if 語句根據(jù)我們所在的 for 循環(huán)的迭代來創(chuàng)建對象playerOne 或playerTwo (i)。不幸的是，java 似乎不會編譯它，因為同一個變量有多個聲明，即使只有其中一個聲明應(yīng)該運行。if(playerNumber == 1){ switch (playerType) { case 1: Human playerOne = new Human(); case 2: Computer playerOne = new Computer(); case 3: Karen playerOne = new Karen(); }}if(playerNumber == 2){ switch (playerType) { case 1: Human playerTwo = new Human(); case 2: Computer playerTwo = new Computer(); case 3: Karen playerTwo = new Karen(); }}OldProgrammer 建議的解決方案如下，效果非常好。但是，我現(xiàn)在在調(diào)用以下方法時遇到問題：人類、計算機和凱倫。嘗試執(zhí)行時出現(xiàn)以下錯誤：System.out.println(player[i].getName() + ", your turn");錯誤消息：Main.java:38: error: cannot find symbol System.out.println(player[i].getName() + ", your turn"); ^symbol: method getName()location: class Player1 error每個類中的方法如下所示：public class Karen extends Player {private String playerName = "Karen";public String getName(){ return playerName;}}
1 回答
白板的微信
TA貢獻1883條經(jīng)驗 獲得超3個贊
感謝老程序員：
更改“人類玩家[i] = new Human();” 到“player[i] = new Human();”等
了解派生類的工作原理非常重要，我很高興這就是它的工作原理。非常感謝你，我希望這可以幫助將來的人。這是我更新的代碼以進行全面說明：
int highestScore = 0;
Player[] player = new Player[2];
Scanner userInput = new Scanner (System.in);
for (int i = 0; i < 2; i++)
{
System.out.println ("Choose a type of player:");
System.out.println ("1: Human");
System.out.println ("2: Computer (Threshold = 20)");
System.out.println ("3. Computer (Custom)");
System.out.print ("? : ");
playerType = Integer.valueOf(userInput.nextLine());
switch (playerType)
{
case 1:
player[i] = new Human();
case 2:
player[i] = new Computer();
case 3:
player[i] = new Karen();
}
}
添加回答
舉報
0/150
提交
取消