正確地拋出你自己的異常（使它不終止你的程序）

Question

我有一些問題拋出我自己的異常。下面是代碼：

class MyException extends Exception { 
    private String message; 

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

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

代碼，其中MyException拋出：

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"); 
    } 

異常通常拋出，該消息被正常打印，但該程序被終止並且「ok」消息不被打印出來。

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

你正在努力嘗試拋出一個新的例外。你不需要這樣做，只需傳遞原文即可。這是更好的做法，因爲FileNotFoundException是一個標準錯誤，所以它是大多數程序員共享的約定。

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(); 
     } 

    } 
}