Different output when Exception Handling Java code is run multiple times

Question

I am aware that many questions about try-catch-finally blocks have been asked in this site. But I have a different doubt. I get different outputs when the code below is run multiple times.

I have a very simple class as follows:

Practice.java

public class Practice {
    public static void main(String []args) {
         System.out.println(getInteger());
    }

    public static int getInteger() {
       try {
           System.out.println("Try");
           throwException();
           return 1;
       } catch(Exception e) {
           System.out.println("Catch Exception");
           e.printStackTrace();
           return 2;
       } finally {
            System.out.println("Finally");
      }
    }

   private static void throwException() throws Exception {
       throw new Exception("my exception");
   }
}

The output when I first run the code above is as follows:

Try
Catch Exception
Finally
2
java.lang.Exception: my exception
    at exceptionHandling.Practice.throwException(Practice.java:22)
    at exceptionHandling.Practice.getInteger(Practice.java:10)
    at exceptionHandling.Practice.main(Practice.java:4)

A different output when I run the code again is given below:

Try
Catch Exception
java.lang.Exception: my exception
    at exceptionHandling.Practice.throwException(Practice.java:22)
    at exceptionHandling.Practice.getInteger(Practice.java:10)
    at exceptionHandling.Practice.main(Practice.java:4)
Finally
2

Can somebody please explain such output?

Jens · Accepted Answer · 2017-03-17 23:02:02Z

5

You use different filehandles. Your output goes to System.out, e.printStackTrace(); writes to System.err which will flush on different times.

edited Mar 17, 2017 at 23:02

answered Dec 19, 2014 at 12:17

Jens

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1 Comment

Ameya Tayade Over a year ago

Thanks. I did not know this. I wanted to find the sequence of execution

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