價值無法解決或不是一個領域

Question

如果this和moving切片具有相同的值，則方法public boolean mergesWith(Tile moving)將返回true 。 但是當我通過執行以下操作檢查它們是否相同時：

if(this.value == temp.value){
    return true;
}

然后它顯示temp.value錯誤，表示無法解析值或不是字段 。

我該如何解決這個問題？

Class TwoNTile ：

package Game2048;

// Concrete implementation of a Tile. TwoNTiles merge with each other
// but only if they have the same value.

public class TwoNTile extends Tile {

    private int value;

    // Create a tile with the given value of n; should be a power of 2
    // though no error checking is done
    public TwoNTile(int n){
        value = n;
    }

    // Returns true if this tile merges with the given tile. "this"
    // (calling tile) is assumed to be the stationary tile while moving
    // is presumed to be the moving tile. TwoNTiles only merge with
    // other TwoNTiles with the same internal value.
    public boolean mergesWith(Tile moving){
        Tile temp = moving;
        if(this.value == temp.value){
            return true;
        }
        else{
            return false;
        }
    }

    // Produce a new tile which is the result of merging this tile with
    // the other. For TwoNTiles, the new Tile will be another TwoNTile
    // and will have the sum of the two merged tiles for its value.
    // Throw a runtime exception with a useful error message if this
    // tile and other cannot be merged.
    public Tile merge(Tile moving){

        return null;
    }

    // Get the score for this tile. The score for TwoNTiles are its face
    // value.
    public int getScore(){

        return -1;
    }

    // Return a string representation of the tile
    public String toString(){

        return "";
    }
}

類瓦片 ：

package Game2048;

// Abstract notion of a game tile.
public abstract class Tile{
  // Returns true if this tile merges with the given tile. 
  public abstract boolean mergesWith(Tile other);

  // Produce a new tile which is the result of merging this tile with
  // the other. May throw an exception if merging is illegal
  public abstract Tile merge(Tile other);

  // Get the score for this tile.
  public abstract int getScore();

  // Return a string representation of the tile
  public abstract String toString();

}

Answer 1

第一 ：你可以這樣做：

public boolean mergesWith(Tile moving){
    return this.value == temp.value;
}

獲得更優雅的解決方案。

第二 ：您需要將value變量添加到Tile類。

public abstract class Tile{
   // ...
   // add a value to Tile
   protected int value; 
   // ...
}

您已擴展Tile並添加了新字段。 那個字段不是Tile的字段，而Tile不包含（看不到）它。

根據以下評論 ：

在Tile聲明值時，不需要再次在TwoNTile聲明它。 你可以制作 Tile對象，而不是使用構造函數。

Tile t = new TwoNTile(...);

是一個有效的Tile對象。 這樣，您就可以實現已經嘗試過的邏輯。

在SO上查看Java中的靜態和動態綁定 ，或者google它 。

Answer 2

您正嘗試從沒有它的抽象類訪問該屬性。

你需要在更高的水平上削減它，例如

TwoNTile temp = (TwoNTile) moving;

我意識到這個抽象類可能有多個擴展，但是你需要將它切換到類層次結構中的最高級別，這樣才能使得類或者它的后代可以回答這個問題。

更新方法：

public boolean mergesWith(Tile moving){
    TwoNTile temp = (TwoNTile) moving;
    if(this.value == temp.value){
        return true;
    }
    else{
        return false;
    }
}

Answer 3

在你的Tile類中，你需要添加類變量value ，如下所示：

package Game2048;

// Abstract notion of a game tile.
public abstract class Tile{

   //need to add instance variable
   int value; 

  // Returns true if this tile merges with the given tile. 
  public abstract boolean mergesWith(Tile other);

  // Produce a new tile which is the result of merging this tile with
  // the other. May throw an exception if merging is illegal
  public abstract Tile merge(Tile other);

  // Get the score for this tile.
  public abstract int getScore();

  // Return a string representation of the tile
  public abstract String toString();

}