關於Java中的一些簡單的異常處理

Question

我想問幾個問題，請參考我在代碼中添加的註釋部分，謝謝。

package test; 

import java.util.InputMismatchException; 
import java.util.Scanner; 

public class Test { 
    /* Task: 
    prompt user to read two integers and display the sum. prompt user to read the number again if the input is incorrect */ 

    public static void main(String[] args) { 
     Scanner input = new Scanner(System.in); 

     boolean accept_a = false; 
     boolean accept_b = false; 
     int a; 
     int b; 

     while (accept_a == false) { 
      try { 
       System.out.print("Input A: "); 
       a = input.nextInt();   /* 1. Let's enter "abc" to trigger the exception handling part first*/ 

       accept_a = true; 
      } catch (InputMismatchException ex) { 
       System.out.println("Input is Wrong"); 
       input.nextLine();     /* 2. I am still not familiar with nextLine() parameter after reading the java manual, would you mind to explain more? All I want to do is "Clear Scanner Buffer" so it wont loop for the println and ask user to input A again, is it a correct way to do it? */ 

      } 
     } 

     while (accept_b == false) { 
      try { 
       System.out.print("Input B: "); 
       b = input.nextInt(); 
       accept_b = true; 
      } catch (InputMismatchException ex) {    /*3. Since this is similar to the above situation, is it possible to reuse the try-catch block to handling b (or even more input like c d e...) exception? */ 

       System.out.println("Input is Wrong"); 
       input.nextLine(); 
      } 

     } 
     System.out.println("The sum is " + (a + b)); /* 4. Why a & b is not found?*/ 

    } 
} 

Answer 1

我仍然不熟悉nextLine（）參數讀數java的手冊後，你會介意解釋嗎？我想要做的只是「清除掃描緩衝區」，因此它不會爲println循環並要求用戶再次輸入A，這是否是正確的方法？

input.nextInt();後使用的input.nextLine();是清除從輸入流中的其餘內容，（至少）的新行字符仍然在緩衝器中，留在緩衝器中的內容將導致input.nextInt();到繼續拋出Exception如果它沒有被清除第一

由於這是與上述類似的情況下，是有可能再利用try-catch塊到處理b（或甚至更多輸入如cd e ...）異常？

你可以，但如果輸入b錯誤會發生什麼？你是否要求用戶重新輸入一個？如果你有100個輸入，並且他們得到最後一個錯誤，會發生什麼？你最好寫一個這樣做的方法，那就是提示用戶輸入一個值並返回該值的方法。

示例...

public int promptForIntValue(String prompt) { 
    int value = -1; 
    boolean accepted = false; 
    do { 
     try { 
      System.out.print(prompt); 
      value = input.nextInt(); 
      accepted = true; 
     } catch (InputMismatchException ex) { 
      System.out.println("Input is Wrong"); 
      input.nextLine(); 
     } 
    } while (!accepted); 
    return value; 
} 

爲什麼一個& b爲沒有發現？

因爲他們還沒有被初始化，編譯器不能確保他們有一個有效的價值...

試着改變它更多的東西一樣。

int a = 0; 
int b = 0; 

Answer 2

1. 是的，沒關係。並將消耗非整數輸入。
2. 是的。如果我們將它提取到一個方法。
3. 因爲編譯器認爲它們可能未被初始化。

讓我們簡化和提取方法，

private static int readInt(String name, Scanner input) { 
    while (true) { 
     try { 
      System.out.printf("Input %s: ", name); 
      return input.nextInt(); 
     } catch (InputMismatchException ex) { 
      System.out.printf("Input %s is Wrong%n", input.nextLine()); 
     } 
    } 
} 

public static void main(String[] args) { 
    Scanner input = new Scanner(System.in); 

    int a = readInt("A", input); 
    int b = readInt("B", input); 
    System.out.println("The sum is " + (a + b)); 
} 

Answer 3

我已經把評論這個問題行。

package test; 

import java.util.InputMismatchException; 
import java.util.Scanner; 

public class Test { 

public static void main(String[] args) { 
    Scanner input = new Scanner(System.in); 

    boolean accept_a = false; 
    boolean accept_b = false; 
    int a=0; 
    int b=0; 

    System.out.print("Input A: "); 

    while (accept_a == false) { 
     try { 

      a = input.nextInt(); // it looks for integer token otherwise exception  

      accept_a = true; 
     } catch (InputMismatchException ex) { 
      System.out.println("Input is Wrong"); 
      input.next(); // Move to next other wise exception // you can use hasNextInt()   

     } 
    } 

    System.out.print("Input B: "); 
    while (accept_b == false) { 
     try { 

      b = input.nextInt(); 
      accept_b = true; 
     } catch (InputMismatchException ex) {  

      System.out.println("Input is Wrong"); 
      input.next(); 
     } 

    } 
    System.out.println("The sum is " + (a + b)); // complier doesn't know wheather they have initialised or not because of try-catch blocks. so explicitly initialised them. 

} 

}

Answer 4

Check out this "nextLine() after nextInt()"

並初始化變量a和b爲零

nextInt（）方法不讀取的最後一個換行符。