正确地抛出你自己的异常（使它不终止你的程序）

Question

我有一些问题抛出我自己的异常。下面是代码：

class MyException extends Exception { 
    private String message; 

    public MyException(String message) { 
     this.message = message; 
    } 

    @Override 
    public String toString() { 
     return "Something went wrong: " + message; 
    } 
} 

代码，其中MyException抛出：

static void expand(String input, String output) throws MyException { 
    try { 
     Scanner sc = new Scanner(new File(input)); 
     //do something with scanner 
    } catch (FileNotFoundException e) { 
     throw new MyException("File not found!"); 
    } 
} 

和主要方法：

public class Encode { 
    public static void main(String[] args) throws MyException { 
     expand("ifiififi.txt", "fjifjif.txt"); 
     System.out.println("ok"); 
    } 

异常通常抛出，该消息被正常打印，但该程序被终止并且“ok”消息不被打印出来。

Exception in thread "main" Something went wrong: File not found! 
at vaje3.Encode.expand(Encode.java:59) 
at vaje3.Encode.main(Encode.java:10) 
Java Result: 1 

Answer 1

你正在努力尝试抛出一个新的例外。你不需要这样做，只需传递原文即可。这是更好的做法，因为FileNotFoundException是一个标准错误，所以它是大多数程序员共享的约定。

public class Encode { 
    static void expand(String input, String output) 
    throws FileNotFoundException 
    { 
     Scanner sc = new Scanner(new File(input));//no try catch here or new exception 
                //just let the original be passed up 
                //via the throw directive 
     //Do more here. 
    } 


    public static void main(String[] args) { 
     try{ 
      expand("ifiififi.txt", "fjifjif.txt"); 
     } catch (FileNotFoundException fnfe){ 
      System.err.println("Warning File not found"); 
      fnfe.printStackTrace(); 
     } 

    } 
}