如何处理符号未找到java中的编译错误？

Question

是给出错误ship1未找到。 我只在符合条件时声明ship1。 其他明智的我已经放置其他条件，重新运行的功能。 这是编译问题，所以我早先被告知...尝试捕捉将无法正常工作。

public static int plantNavy(BattleBoard myBattle, int counter) { 
    System.out.println("Im innnnn"); 
    if (counter == 0) { 
     System.out.println("\nPlacing Large Ship"); 
    } 
    else if (counter == 1) { 
     System.out.println("Placing Medium Ship"); 
    } 
    else if (counter == 2) { 
     System.out.println("Placing Medium Ship"); 
    } 
    else if (counter == 3) { 
     System.out.println("Placing Small Ship"); 
    } 
    else if (counter == 4) { 
     System.out.println("Placing Small Ship"); 
    } 

    System.out.println("Enter 0 to place ship horizontally"); 
    System.out.println("Enter 1 to place ship vertically"); 

    String align = shipAlignment.nextLine(); 
    if (align.length() > 1) { 
     System.out.println("Inappropriate value entered. Please enter again"); 
     plantNavy(myBattle,counter); 
    } 

    if (align.charAt(0) - 48 == 0 || align.charAt(0) - 48 == 1) { 
     if (align.charAt(0) - 48 == 0) { 
      if (counter == 0) { 
       BattleShip ship1 = new LargeShip(false); 
      } 
      else if (counter == 1) { 
       BattleShip ship1 = new MediumShip(false); 
      } 
      else if (counter == 2) { 
       BattleShip ship1 = new MediumShip(false); 
      } 
      else if (counter == 3) { 
       BattleShip ship1 = new SmallShip(false); 
      } 
      else if (counter == 4) { 
       BattleShip ship1 = new SmallShip(false); 
      } 
     } 
     if (align.charAt(0) - 48 == 1) { 
      if (counter == 0) { 
       BattleShip ship1 = new LargeShip(true); 
      } 
      else if (counter == 1) { 
       BattleShip ship1 = new MediumShip(true); 
      } 
      else if (counter == 2) { 
       BattleShip ship1 = new MediumShip(true); 
      } 
      else if (counter == 3) { 
       BattleShip ship1 = new SmallShip(true); 
      } 
      else if (counter == 4) { 
       BattleShip ship1 = new SmallShip(true); 
      } 
     } 
    } 
    else { 
     System.out.println("Inappropriate value entered"); 
     counter=plantNavy(myBattle,counter); 
    } 

     System.out.println("Enter Ship Placing position"); 
     String shipPos = shipPlace.next(); 
     if (shipPos.length() > 3 || shipPos.length() < 2) { 
      System.out.println("Inappropriate target. Please enter again"); 
      counter = plantNavy(myBattle,counter); 
     } 
     else if ((int) (shipPos.charAt(1))-48 < 1 || (int) shipPos.charAt(1)-48 > 10) { 
      System.out.println("Inappropriate target. Please enter again"); 
      counter = plantNavy(myBattle,counter); 
     } 

     else if ((int) (shipPos.charAt(0)) < 65 || (int) shipPos.charAt(0)> 74) { 
      System.out.println("Inappropriate target. Please enter again"); 
      counter = plantNavy(myBattle,counter); 
     } 

     int x_pos; 
     int y_pos; 

     if (shipPos.length() == 3) { 
      shipPos = shipPos.charAt(0) + "10"; 
     } 
     if (shipPos.length() == 2) { 
      x_pos = (int) (shipPos.charAt(1))-49; 
     } 
     else { 
      x_pos = 9; 
     } 
     y_pos = (int) (shipPos.charAt(0))-65; 

     System.out.println(x_pos); 
     System.out.println(y_pos); 

     boolean plantCor = myBattle.addShip(ship1,x_pos,y_pos); 

     if (plantCor == true) { 
      System.out.println(myBattle.printActualBoard()); 
      counter++; 
      return counter; 
     } 

     if (plantCor == false) { 
      System.out.println("Incorrect Placement. Place Again in empty area."); 
      counter = plantNavy(myBattle,counter); 
     } 
    } 

Answer 1

在你的方法顶部声明战舰SHIP1 = NULL变量并使用它：

public static int plantNavy(BattleBoard myBattle, int counter) { 
    BattleShip ship1 = null; 

在if块，只是分配给它不要再声明变量。即，做

if (counter == 0) { 
    ship1 = new LargeShip(false); 
} 

不

if (counter == 0) { 
    BattleShip ship1 = new LargeShip(false); // note the difference! 
} 

以后你可以检查它是否是空，看是否SHIP1已创建。

if (ship1 == null) { 
    // then no ship1 has been created yet. 
} 

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编辑
幽州的评论：

@HovercraftFullOfEels我尝试这种技术，但它给缺少return语句的错误。只有在船舶成功放置后才能返回。

然后给它必要的返回语句。你的代码有一堆返回嵌套在if语句中。如果if语句都不是真的，那么方法有可能结束而不返回任何内容，这是不允许的。

一种解决方案是在方法底部添加一个默认返回语句，但如果这是我的代码，我会查看整个方法并重新编写它，将其重构为几个更小的更简单的方法，用户交互出这个方法。

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编辑2
我看到的另一个问题是，你在一个危险的和不必要的方式使用递归：

if (shipPos.length() > 3 || shipPos.length() < 2) { 
    System.out.println("Inappropriate target. Please enter again"); 
    counter = plantNavy(myBattle, counter); 
    } 

需要注意的是如果输入错误的输入，你打电话该方法再次在相同的方法内，但没有意识到，即使该方法将再次调用，在递归调用返回后，原始方法仍将运行完成与错误输入。我的建议，避免在这里递归，再次重构和简化你的代码将帮助你做到这一点。

取而代之的是，获取用户输入的一个单独的方法，验证输入，因为它正在被获取（一个简单的do-while循环就足够了），然后一旦获得了所有必要的放置船的输入，调用你的方法放置一艘而没有其中有任何用户输入。

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编辑3
下一步，你想摆脱那些神奇的数字，你正在使用的，使你的代码熊理解和调试。