Question

所以，我搜索了谷歌和stackoverflow一点，但我似乎无法找到这个问题的答案。

我有两种方法。第一种方法getAge只是从用户那里得到一个整数作为输入。然后，它意味着将该输入传递给verifyAge，后者确保它在正确的范围内。

然而;如果他们应该输入任何不是整数的东西，它应该显示一条消息并再次调用getAge，以重新启动输入过程。我有一个try-catch设置，但它仍然可以回到JVM。根据另一篇文章的回答;我正在做的是正确的。但它似乎仍然没有奏效。所以这是我尝试运行它时遇到的错误：

Please enter your age: notint
Exception in thread "main" java.util.InputMismatchException
 at java.util.Scanner.throwFor(Scanner.java:864)
 at java.util.Scanner.next(Scanner.java:1485)
 at java.util.Scanner.nextInt(Scanner.java:2117)
 at java.util.Scanner.nextInt(Scanner.java:2076)
 at Ch2ProgLabWilson.getAge(Ch2ProgLabWilson.java:22)
 at Ch2ProgLabWilson.main(Ch2ProgLabWilson.java:15)

我写的：

import java.util.* ;
import java.util.Scanner;

public class Ch2ProgLabWilson {

 public static void main(String[] args) {

      getAge();  
 }

 public static int getAge()
 {
  Scanner keyboard = new Scanner(System.in);
  System.out.print("Please enter your age: ");
  int a = keyboard.nextInt();
  verifyAge(a);

     try
     {
        getAge();
     }    
     catch (InputMismatchException e)
        {
           System.out.println("You may only enter integers as an age. Try again.");
           getAge();
        }

    return a;
   }

 // 
 public static boolean verifyAge (int a)
     {
        if (a >= 0 && a <= 122)
        {
           System.out.println("The age you entered, " + a + ", is valid.");
           return true;
        }
        else
        {
           System.out.println("The age must be from 0 to 122, cannot be negative, and has to be an integer.");
           getAge();
           return false;
        }   
     }

 }

Answer 1

int a = keyboard.nextInt();抛出了异常，它位于try catch块之外。将调用放在try块中的int a = keyboard.nextInt();。

您的代码还有其他问题：

verifyAge()会返回一个从未使用过的boolean。

你的getAge()方法是递归的，假设用户输入一个数字，它就会循环 - 这是你想要的吗？

<强>更新

public static int getAge(){
    Scanner keyboard = new Scanner(System.in);
    System.out.print("Please enter your age: ");
    int age = -1;

    while(!verifyAge(age)){ // will loop until there's a valid age
        try{
            age = scanner.nextInt();
        catch (InputMismatchException e){
            System.out.println("You may only enter integers as an age. Try again.");
        }
    }

    return age; // your current code doesn't do anything with this return value
}

public static boolean verifyAge (int a){ // would be better named isValidAge()
    if (a >= 0 && a <= 122){
        System.out.println("The age you entered, " + a + ", is valid.");
        return true;
    }else{ // no need to call getAge() here
       System.out.println("The age must be from 0 to 122, cannot be negative, and has to be an integer.");
       return false;
    }   
 }

Answer 2

代码不在try块

内

  Scanner keyboard = new Scanner(System.in);
  System.out.print("Please enter your age: ");
  int a = keyboard.nextInt();
  verifyAge(a);

  try
  {

所以它不会被抓住。

Answer 3

查看实际抛出异常的代码行。如果它不在try/catch块中，则不会被捕获。

请改为尝试：

try
{
    int a = keyboard.nextInt();
    verifyAge(a);
}    
catch (InputMismatchException e)
{
    System.out.println("You may only enter integers as an age. Try again.");
    getAge();
}

Catch没有捕获InputMismatchException？

3 个答案: