Java Exception Handling understanding issue

Question

I'm not able to understand this program. I expect it to output "Hello World", but instead it prints only the "World". I thought that first the try block would execute, printing "Hello" and " ", then afterward when it encounters a 1/0, it would throw an ArithmeticException. The exception would be caught by catch block, then "World" would be printed.

The program is as follows.

 import java.util.*;
 class exception{
     public static void main(String args[]) 
     {
         try
         {
             System.out.println("Hello"+" "+1/0);
         } 
         catch(ArithmeticException e) 
         {
             System.out.println("World");
         }
     }
 }

Why do you think it would print "Hello" before evaluating 1/0? — user2357112
– user2357112, Commented Jun 29, 2016 at 18:22
It has to interpret the value of "Hello" + " " + 1/0 before it prints anything. "Hello" isn't printed because you aren't saying to print just "Hello", rather "Hello" plus something that causes the exception. — Chris Sprague
– Chris Sprague, Commented Jun 29, 2016 at 18:23

Chris Sprague · Accepted Answer · 2016-06-29 18:35:22Z

5

The exception is thrown before calling the println function. The argument value has to be calculated before the function call.

In order for your program to achieve the results you expect, you would edit the code in the try block as follows:

     try
     {
         // this will work and execute before evaluating 1/0
         System.out.print("Hello ");
         // this will throw the exception
         System.out.print(1/0);
     } 
     catch(ArithmeticException e) 
     {
         System.out.println("World");
     }

edited Jun 29, 2016 at 18:35

Chris Sprague

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answered Jun 29, 2016 at 18:23

Frank Puffer

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Comments

user6499716 · Accepted Answer · 2016-06-29 18:29:51Z

1

First "Hello"+" "+1/0 will be evaluated. And then passed as an argument to System.out.println(...). That's why an exception is thrown before System.out.println(...) would have been called.

answered Jun 29, 2016 at 18:29

user6499716

Comments

Doug · Accepted Answer · 2016-06-29 18:29:40Z

1

It does not simply scan the "words" left to right. Everything inside the ( ) needs to be evaluated successfully, and if it is then it gets printed.

It looks at "Hello" and it's fine. Next it looks at 1/0 and creates error.

In the event that the math evaluated successfully it would attempt to concatenate "Hello" and the result. And if that was successful, then it would be printed.

answered Jun 29, 2016 at 18:29

Doug

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1 Comment

Sandeep Roy Over a year ago

Understood sir. really appreciate.

Junaid Aslam · Accepted Answer · 2021-12-27 18:14:34Z

1

It will checked statement by statement. So, it checks for whole println argument. But, it has an exception, so goes for executing catch block.

NOTE: If it execute first half statement and than checks for Exception, no need remains for try-catch block.

answered Dec 27, 2021 at 18:14

Junaid Aslam

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Comments

chandu raju · Accepted Answer · 2024-03-19 18:45:55Z

1

Arguments that are passed in println function/method will be checked first followed by calling of println.So , exception will rise before calling println.Since exception is raised control will go to catch and only "World" will be printed

edited Mar 19, 2024 at 18:45

answered Dec 27, 2021 at 18:08

chandu raju

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