如何處理符號未找到java中的編譯錯誤？

Question

是給出錯誤ship1未找到。 我只在符合條件時聲明ship1。 其他明智的我已經放置其他條件，重新運行的功能。 這是編譯問題，所以我早先被告知...嘗試捕捉將無法正常工作。

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

在你的方法頂部聲明戰艦SHIP1 = NULL變量並使用它：

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

在if塊，只是分配給它不要再聲明變量。即，做

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

不

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

以後你可以檢查它是否是空，看是否SHIP1已創建。

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

---

編輯
幽州的評論：

@HovercraftFullOfEels我嘗試這種技術，但它給缺少return語句的錯誤。只有在船舶成功放置後才能返回。

然後給它必要的返回語句。你的代碼有一堆返回嵌套在if語句中。如果if語句都不是真的，那麼方法有可能結束而不返回任何內容，這是不允許的。

一種解決方案是在方法底部添加一個默認返回語句，但如果這是我的代碼，我會查看整個方法並重新編寫它，將其重構爲幾個更小的更簡單的方法，用戶交互出這個方法。

---

編輯2
我看到的另一個問題是，你在一個危險的和不必要的方式使用遞歸：

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

需要注意的是如果輸入錯誤的輸入，你打電話該方法再次在相同的方法內，但沒有意識到，即使該方法將再次調用，在遞歸調用返回後，原始方法仍將運行完成與錯誤輸入。我的建議，避免在這裏遞歸，再次重構和簡化你的代碼將幫助你做到這一點。

取而代之的是，獲取用戶輸入的一個單獨的方法，驗證輸入，因爲它正在被獲取（一個簡單的do-while循環就足夠了），然後一旦獲得了所有必要的放置船的輸入，調用你的方法放置一艘而沒有其中有任何用戶輸入。

---

編輯3
下一步，你想擺脫那些神奇的數字，你正在使用的，使你的代碼熊理解和調試。