关于Java中的一些简单的异常处理

Question

我想问几个问题，请参考我在代码中添加的注释部分，谢谢。

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（）参数读数java的手册后，你会介意解释吗？我想要做的只是“清除扫描缓冲区”，因此它不会为println循环并要求用户再次输入A，这是否是正确的方法？

input.nextInt();后使用的input.nextLine();是清除从输入流中的其余内容，（至少）的新行字符仍然在缓冲器中，留在缓冲器中的内容将导致input.nextInt();到继续抛出Exception如果它没有被清除第一

由于这是与上述类似的情况下，是有可能再利用try-catch块到处理b（或甚至更多输入如cd e ...）异常？

你可以，但如果输入b错误会发生什么？你是否要求用户重新输入一个？如果你有100个输入，并且他们得到最后一个错误，会发生什么？你最好写一个这样做的方法，那就是提示用户输入一个值并返回该值的方法。

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; 
} 

为什么一个& b为没有发现？

因为他们还没有被初始化，编译器不能确保他们有一个有效的价值...

试着改变它更多的东西一样。

int a = 0; 
int b = 0; 

Answer 2

1. 是的，没关系。并将消耗非整数输入。
2. 是的。如果我们将它提取到一个方法。
3. 因为编译器认为它们可能未被初始化。

让我们简化和提取方法，

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

我已经把评论这个问题行。

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()"

并初始化变量a和b为零

nextInt（）方法不读取的最后一个换行符。