Question

我正在编写一个创建游戏板的程序，并使用超类和子类来生成游戏板。本质上，我想使用一个arrayList来存储从Cell子类的方法返回的值。但我无法弄清楚该怎么做。我在BoardGame类中写的内容给了我一个错误，我想不出另一种方法来做到这一点。 GameBoard

import java.util.ArrayList;


public class BoardGame {



    private Die theDie;



    BoardGame(){





    }

    void buildBoard(){

        ArrayList<Player> player;




        Cell c = new Cell();

        ArrayList<Cell>cells = new ArrayList<Cell>();

        cells.add(c.landOn());

    }

    void runSimulation(){

    }


     int takeTurn(String n){

         //since we don't know what we are returning yet
         return 0;


    }

     String toString(String n){


         //Since we aren't returning anything yet
         return "Nothing";


     }
}






public class Cell {

    static int landOn(){


        //Since we do not know what we are returning
        return 0;
    }

    public String toString(){



        //Since we do not know what we are returning
        return "Blank";
    }

}




public class Star extends Cell {

    int landOn(){


        //since we do not know what we are returning 
        return 5;
    }

    public String toString(){


        //since we do not know what we are returning
        return "Star: +5";
    }

}



public class Lightning extends Cell {

    int landOn(){


        //since we do not know what we are returning 
        return -5;
    }

    public String toString(){


        //since we do not know what we are returning
        return "Lightning: -5";
    }

}



public class X extends Cell {

    int landOn(){


        //since we do not know what we are returning 
        return -10;
    }

    public String toString(){


        //since we do not know what we are returning
        return "X: -10";
    }

}



public class Smiley extends Cell {

    int landOn(){


        //since we do not know what we are returning 
        return 10;
    }

    public String toString(){


        //since we do not know what we are returning
        return "Smiley: +10";
    }

}

Answer 1

在代码中使用ArrayList<Cell>将无效，因为c.landOn()会返回int。使用Integer返回类型ArrayList<Integer>cells将解决该特定问题。那是你看到的错误吗？

PS：还记得你可以说ArrayList<Cell>cells = new ArrayList<>();而不是ArrayList<Cell>cells = new ArrayList<Cell>();

Answer 2

您的ArrayList被声明为Cell 方法的返回类型为int。您的ArrayList无法存储除Cell

类型的实例以外的任何内容

要存储所需方法的所有返回类型，需要ArrayList类型的Integer

使用int无法正常工作int是原始的。

解决方案：

ArrayList<Integer> cells = new ArrayList<>();

Answer 3

你可以这样做。

BoardGame课程：

public class BoardGame {
    public BoardGame(){
        //Get finalTreats from Cell class
        int finalTreats = this.cell.finalTreats;
        //Add finalTreats to cells ArrayList
        cells.add(finalTreats);
    }

    //Get Cell class
    Cell cell = new Cell();

    //Initialize ArrayList
    ArrayList<Integer> cells = new ArrayList<>();
}

细胞类：

public class Cell {
    public int finalTreats;
    public int landOn(int treats){
        //Test for what it landed on based on treat amount
        switch(treats){
            //X
            case -10: finalTreats = -10;
            //Lightning
            case -5: finalTreats = -5;
            //Star
            case 5: finalTreats = 5;
            //Smiley face
            case 10: finalTreats = 10;

            //Blank
            default: return 0;
        }
    }
}

星级（例如）：

public class Star extends Cell{
    public Star() {
        this.landOn(5);
    }
}

使其他类如Lighting和Smiley Face与Star类相同，但根据需要更改处理的值。

希望它有所帮助。

Answer 4

您也可以这样做：

public abstract class Cell{
...
}
public class EmptyCell extends Cell{
...
}
public class Win5Treats extends Cell{
...
}
public class Lose5Treats extends Cell{
...
}
public class Win10Treats extends Cell{
...
}
public class Lose10Treats extends Cell{
...
}    
public class Start extends Cell{
...
}
public class End extends Cell{
...
}

你的阵列就是这样的：

ArrayList<Cell> cells = new ArrayList<Cell>();

在ArrayList中存储方法返回值

4 个答案: