Enthuware

Posted: **Mon Sep 24, 2012 9:01 am**

I don't see the args[2] reference in the code.

Posted: **Mon Sep 24, 2012 9:09 am**

int i = Integer.parseInt(args[1]);   
System.out.println(args[i]);

args[1] is 2
so i is 2
therefore, args is args[2]

HTH,
Paul.

Posted: **Mon Sep 24, 2012 9:29 am**

I should pay more attention

.

Posted: **Wed Feb 13, 2013 10:16 am**

Yeah, this one got me also

Posted: **Tue Jun 18, 2013 11:00 am**

I have a basic question,
why is the method name just parseInt() instead of parseInteger()?

Please reply,

Posted: **Tue Jun 18, 2013 11:13 am**

You have to ask Java designers that

Posted: **Wed Apr 15, 2015 7:18 am**

If I run the code with 4 and 5 I also get the exception, in this case shouldn't args[0] = 4 and args[1]=5?

Posted: **Wed Apr 15, 2015 10:32 am**

peckw1 wrote:If I run the code with 4 and 5 I also get the exception, in this case shouldn't args[0] = 4 and args[1]=5?

Yes, args[0] = 4 and args[1]=5. But I am not really sure what is your point.
Paul.

Posted: **Sat Jul 11, 2020 4:31 pm**

System.out.println(i);//2
System.out.println(args); //i=2, so, args[2] --> throw an exception
Yes, this one got me also....

About Question enthuware.ocajp.i.v7.2.1242 :