java - 值无法解析或不是字段

Question

方法 public boolean mergesWith(Tile moving) 返回 true 如果 this 和 moving 瓷砖具有相同的值(value)。但是当我通过执行以下操作检查它们是否相同时:

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

然后它向我显示关于 temp.value 的错误，说 value cannot be resolved or is not a field。

我该如何解决？

TwoNTile类:

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 "";
    }
}

瓷砖类:

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

首先:你可以这样做:

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

以获得更优雅的解决方案。

其次:您需要将value 变量添加到Tile 类。

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

您已经扩展了 Tile 并添加了新字段。该字段不是 Tile 的字段，Tile 不包含(看不到)它。

基于以下评论:

当您在Tile 中声明值时，您不需要在TwoNTile 中再次声明它。 您可以制作 Tile 对象，只是不使用构造函数。

Tile t = new TwoNTile(...);

是一个有效的 Tile 对象。这样，您就可以实现您已经尝试使用的逻辑。

看static and dynamic binding在 SO 上的 Java 中，或 google it .