Try/Catch變量無法初始化

Question

我遇到了try/catch塊的問題。現在，我的代碼在某種程度上起作用，但是，例如，當輸入一個無效的數字（「a」）時，我打印出該程序（「請輸入一個有效的數字：」）;但是，我有一個方法可以在輸入的索引處獲取對象，並說明局部變量可能尚未初始化。

do { 
     int firstNum; 
     int secondNum; 
     int thirdNum; 
     System.out 
       .println("Please enter three of the numbers you see on the left of the shape:"); 
     try { 

      firstNum = scan.nextInt(); 
      secondNum = scan.nextInt(); 
      thirdNum = scan.nextInt(); 

      Card card = Deck.randomizedCards.get(firstNum); 
      Card card1 = Deck.randomizedCards.get(secondNum); 
      Card card2 = Deck.randomizedCards.get(thirdNum); 
      if (Deck.isSet(card, card1, card2) == true) { 
       JOptionPane.showMessageDialog(panel, "You found a set!"); 
       Deck.completedSets.add(card); 
       Deck.completedSets.add(card1); 
       Deck.completedSets.add(card2); 

      } else { 
       JOptionPane.showMessageDialog(panel, 
         "You didn't find a set"); 
      } 
      break; 
     } catch (InputMismatchException name) { 
      System.out 
        .println("You have entered an invalid number, please enter a number:"); 

     } catch (ArrayIndexOutOfBoundsException name) { 
      System.out 
        .println("You have entered an invalid number, please enter a number:"); 

     } 

    } while (true); 

Answer 1

）的主要問題就出在這代碼片段

try{ 
    firstNum= scan.nextInt(); 
. 
. 
. 
catch (InputMismatchException name) { 
    System.out.println("You have entered an invalid number, please enter a number (Range 1-11): "); 

} 

當輸入異常被拋出違規輸入被保留在流所以當執行到達scan.nextInt（它再次得到了相同的不良輸入，因此拋出相同的異常等。將catch塊更改爲：

catch (InputMismatchException name) { 
    String malformedGarbage = scan.Next();//processes the bad input so the scanner can move on. 
    System.out.println("You have entered an invalid number, please enter a number (Range 1-11): "); 

Answer 2

由於在try語句中的分配可能不會發生你有你的變量firstNum，secondNum，thirdNum留下潛在unititialized。

您可以通過分配默認值或將邏輯移入相同的try語句來規避該情況。

Answer 3

編譯器（正確）警告您，例如，如果行secondNum = scan.nextInt();引發異常，則變量（如thirdNum）可能未被初始化。

爲了規避這個問題，您應該爲所有變量指定默認值，或者在拋出異常時使該方法返回。

Answer 4

int firstNum; 
int secondNum; 
int thirdNum; 
do { 
    System.out.println("Please enter three of the numbers you see on the left of the shape:"); 
    try { 
     firstNum = scan.nextInt(); 
     secondNum = scan.nextInt(); 
     thirdNum = scan.nextInt(); 
     Card card = Deck.randomizedCards.get(firstNum); 
     Card card1 = Deck.randomizedCards.get(secondNum); 
     Card card2 = Deck.randomizedCards.get(thirdNum); 
     if (Deck.isSet(card, card1, card2)) { 
      JOptionPane.showMessageDialog(panel, "You found a set!"); 
      Deck.completedSets.add(card); 
      Deck.completedSets.add(card1); 
      Deck.completedSets.add(card2); 
      break; 
     } else { 
      JOptionPane.showMessageDialog(panel, "You didn't find a set"); 
     } 
    } catch (InputMismatchException name) { 
     System.out 
       .println("You have entered an invalid number, please enter a number (Range 1-11): "); 

    } catch (ArrayIndexOutOfBoundsException name) { 
     System.out 
       .println("You have entered an invalid number, please enter a number(Range 1-11): "); 
    } 
} while (true);